THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS

Excerpts

THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
This theorem asserts that if we have a convergent series of positive terms, $u_{0} + u_{1} + u_{2} + \dots$ say, and form any other series
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The tests derived from comparison with it are therefore naturally very crude, and much more delicate tests are often wanted.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
This hardly requires proof, for $v_{n}^{1/n} \geq 1$ involves $v_{n} \geq 1$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
If $0 < a < b < 1$, then the series $a + b + a^{2} + b^{2} + a^{3} + \dots$ is convergent.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The explanation is to be found in a closer consideration of the relation between $x$ and $y$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
This example shows that the condition that $\phi_{n}$ should tend *steadily* to zero is essential to the truth of the theorem.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Dirichlet’s Theorem ([§]169) shows that the terms of a series of positive terms may be rearranged in any way without affecting its sum.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
if $\sum u_{n}$ is absolutely convergent then it is convergent; so are the series formed by its positive and negative terms taken separately; and the sum of the series is equal to the sum of the positive terms plus the sum of the negative terms.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
There is another test, due to Abel, which, though of less frequent application than Dirichlet’s, is sometimes useful.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
We can now extend this result to all cases in which $\sum u_{n}$ and $\sum v_{n}$ are *absolutely* convergent; for our proof was merely a simple application of Dirichlet’s Theorem, which we have already extended to all absolutely convergent series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Deduce that *if $\sum c_{n}$ is convergent then its sum is $AB$*.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
In case (3) the circle is called the **of convergence** and its radius the **of convergence** of the power series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Moreover, in inferring the convergence or divergence of $\sum v_{n}$ by means of one of these tests, it is sufficient to know that the test is satisfied for *sufficiently large* values of $n$, *i.e.* for all values of $n$ greater than a definite value $n_{0}$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
None the less d’Alembert’s test is very useful in practice, because when $v_{n}$ is a complicated function $v_{n+1}/v_{n}$ is often much less complicated and so easier to work with.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Let $s$ be the sum of the series of $u'$s. Then the sum of any number of terms, selected from the $u'$s, is not greater than $s$. But every $v$ is a $u$, and therefore the sum of any number of terms selected from the $v'$s is not greater than $s$. Hence $\sum v_{n}$ is convergent, and its sum $t$ is not greater than $s$. But we can show in exactly the same way that $s \leq t$. Thus $s = t$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
This theorem was discovered by Abel but forgotten, and rediscovered by Pringsheim.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
*The converse of Abel’s theorem is not true*, *i.e.* it is not true that, if $u_{n}$ decreases with $n$ and $\lim nu_{n} = 0$, then $\sum u_{n}$ is convergent.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The fact is that the geometric series, by comparison with which the tests of [§]168 were obtained, is not only convergent but *very rapidly* convergent, far more rapidly than is necessary in order to ensure convergence. The tests derived from comparison with it are therefore naturally very crude, and much more delicate tests are often wanted.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
But this is far from being the case; if only we go far enough into the sequences we shall find the terms of the first sequence very much the smaller.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The second of the two tests mentioned in [§]172 is as follows: Resultif $u_{n} = \phi(n)$ is a decreasing function of $n$, then the series $\sum \phi(n)$ is convergent or divergent according as $\sum 2^{n}\phi(2^{n})$ is convergent or divergent. Result
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
For our present purposes the field of application of this test is practically the same as that of the Integral Test. It enables us to discuss the series $\sum n^{-s}$ with equal ease. For $\sum n^{-s}$ will converge or diverge according as $\sum 2^{n}2^{-ns}$ converges or diverges, *i.e.* according as $s > 1$ or $s \leq 1$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Of course the reader will not be puzzled by the use of the term *infinite integral* to denote something which has a definite value such as $2$ or $\frac{1}{2}\pi$. The distinction between an infinite integral and a finite integral is similar to that between an infinite series and a finite series: no one supposes that an infinite series is necessarily divergent.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The integral $\ds\int_{a}^{x} \phi(t)\, dt$ was defined in [§§]156 and 157 as a *simple* limit, *i.e.* the limit of a certain finite sum. The infinite integral is therefore *the limit of a limit*, or what is known as a *repeated* limit. The notion of the infinite integral is in fact essentially more complex than that of the finite integral, of which it is a development.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
There is one fundamental property of a convergent infinite series in regard to which the analogy between infinite series and infinite integrals breaks down. If $\sum \phi(n)$ is convergent then $\phi(n) \to 0$; but it is *not* always true, even when $\phi(x)$ is always positive, that if $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent then $\phi(x) \to 0$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
It often happens that the subject of integration has a discontinuity which is due simply to a failure in its definition at a particular point in the range of integration, and can be removed by attaching a particular value to it at that point. In this case it is usual to suppose the definition of the subject of integration completed in this way.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
An integral in which the subject of integration tends to $\infty$ or to $-\infty$ as $x$ tends to some value or values included in the range of integration will be called an *infinite integral of the second kind*: the *first kind* of infinite integrals being the class discussed in [§§]177 *et seq.*
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Some care has occasionally to be exercised in applying the rule for transformation by substitution. The following example affords a good illustration of this.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
The reader should carefully guard himself against supposing that the statement ‘an absolutely convergent series is convergent’ is a mere tautology. When we say that $\sum u_{n}$ is ‘absolutely convergent’ we do *not* assert directly that $\sum u_{n}$ is convergent: we assert the convergence of *another* series $\sum |u_{n}|$, and it is by no means evident *a priori* that this precludes oscillation on the part of $\sum u_{n}$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
In the first place we note that, if $\sum u_{n}$ is conditionally convergent, then the series $\sum v_{n}$, $\sum w_{n}$ of [§]184 must both diverge to $\infty$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
In the first instance *there are no comparison tests for convergence of conditionally convergent series*.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
We shall see shortly that the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ is convergent. But the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$ is divergent, although each of its terms is numerically less than the corresponding term of the former series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
It can indeed be proved that a conditionally convergent series can always be so rearranged as to converge to any sum whatever, or to diverge to $\infty$ or to $-\infty$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
This theorem may be stated as follows: Resulta convergent series remains convergent if we multiply its terms by any sequence of positive and decreasing factors. Result
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
It is obvious that *an absolutely convergent series is convergent*, since its real and imaginary parts converge separately.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
For $\lim a_{n}z_{1}^{n} = 0$, since $\sum a_{n}z_{1}^{n}$ is convergent, and therefore we can certainly find a constant $K$ such that $|a_{n}z_{1}^{n}| < K$ for all values of $n$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
In other words, if the series converges at $P$ *then it converges absolutely at all points nearer to the origin than $P$*.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
It should be observed that this general result gives absolutely no information about the behaviour of the series *on* the circle of convergence. The examples which follow show that as a matter of fact there are very diverse possibilities as to this.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Thus the logarithmic series converges at all points of its circle of convergence except the point $z = -1$.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
It shows that *the same function $f(z)$ cannot be represented by two different power series*.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
Such equations may be solved by a method which will be sufficiently explained by an example.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
A player tossing a coin is to score one point for every head he turns up and two for every tail, and is to play on until his score reaches or passes a total $n$.

Equations

THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{m} + u_{m+1} + \dots + u_{n} = \sum_{m}^{n} \phi(\nu)
The sum of the terms from u_m to u_n is written as a sum of phi(nu) from m to n, which is the book's shorthand notation for partial sums. Flag: the right-hand summand is printed as phi(nu) although the left side is in u; this is the book's own notation and is reported as printed.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{0} + u_{1} + \dots + u_{n} < K
A series of positive terms is convergent exactly when there is a number K that bounds all its partial sums from above.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n} \leq Ku_{n}
If v_n is at most K times u_n for all sufficiently large n and the series of u_n converges, then the series of v_n converges (Theorem C).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum v_{n} \leq K \sum u_{n}
Under the comparison hypothesis, the sum of the v_n is at most K times the sum of the u_n (Theorem C).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n} \leq Kr^{n}
If v_n is at most K r^n for all sufficiently large n, with r less than 1, the series of v_n is convergent, by comparison with the geometric series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n}^{1/n} \leq r
Cauchy's test: the series of positive terms v_n converges if the nth root of v_n is at most r, with r less than 1, for all sufficiently large n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n}^{1/n} \geq 1
The divergence form of Cauchy's test: the series of v_n diverges if the nth root of v_n is at least 1 for an infinity of values of n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n+1}/v_{n} \leq r
d'Alembert's ratio test: the series of v_n converges if the ratio of successive terms is at most r, with r less than 1, for all sufficiently large n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n+1}/v_{n} \geq r \geq 1
If the ratio of successive terms is at least r, with r at least 1, for all (or all sufficiently large) n, the series of v_n diverges. The book notes that the same bound holding only for an infinity of n does not suffice.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{n+1}/v_{n} \to l
The footnote states that if the ratio v_{n+1}/v_n tends to l, then v_n^{1/n} also tends to l (proof deferred to a later chapter); the converse is false.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{0} v_{0} + (u_{1} v_{0} + u_{0} v_{1}) + (u_{2} v_{0} + u_{1} v_{1} + u_{0} v_{2}) + \dots
The product of two convergent series of positive terms, with sums s and t, is the convergent series whose terms are grouped by total suffix; its sum is st.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
(u_{0} + u_{1} + \dots + u_{n})(v_{0} + v_{1} + \dots + v_{n})
The sum of the first n+1 groups in the product arrangement equals the product of the two partial sums, and so tends to st.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = \phi(n)
The term u_n of the series is written as phi(n), the value of a decreasing continuous function phi(x) at x = n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n+1} \leq u_{n}
The assumed condition that the terms decrease steadily: each term is at most the one before it, for all or all sufficiently large n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim nu_{n} = 0
Abel's theorem: if the series of u_n is convergent with positive decreasing terms, then n u_n tends to zero. It is one-sided: it gives divergence tests but not convergence.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\phi(\nu - 1) \geq \phi(x) \geq \phi(\nu)
For x between nu-1 and nu, phi(x) lies between phi(nu-1) and phi(nu), because phi is steadily decreasing.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{\nu} = \phi(\nu - 1) - \int_{\nu-1}^{\nu} \phi(x)\, dx
v_nu is defined as the difference between phi(nu-1) and the integral of phi over the interval from nu-1 to nu, used in the integral test.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
0 \leq v_{\nu} \leq \phi(\nu - 1) - \phi(\nu)
The auxiliary term v_nu is non-negative and at most phi(nu-1) minus phi(nu).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{2} + v_{3} + \dots + v_{n} \leq \phi(1) - \phi(n) \leq \phi(1)
The partial sums of the auxiliary series are bounded by phi(1), so that series converges.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\Phi(\xi) = \int_{1}^{\xi} \phi(x)\, dx
Phi(xi) is defined as the integral of phi from 1 to xi; it is continuous and steadily increasing in xi.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum_{1}^{n-1} \phi(\nu) - \int_{1}^{n} \phi(x)\, dx
The difference between the partial sums of phi(nu) and the integral up to n tends to a positive limit as n tends to infinity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
v_{\nu} < \phi(\nu - 1) - \phi(\nu)
Remark: the strict form of the bound for the auxiliary term holds unless phi is constant on the interval, so the sum is strictly less than phi(1) + l.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\phi(1) + \phi(2) + \dots
Maclaurin's (or Cauchy's) integral test: for a positive continuous decreasing phi, the series phi(1)+phi(2)+... converges if and only if the integral of phi from 1 to xi tends to a limit as xi tends to infinity; its sum is then at most phi(1) + l.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}} = \frac{\xi^{1-s} - 1}{1 - s}
For s not equal to 1, the integral of x^(-s) from 1 to xi equals (xi^(1-s) - 1)/(1 - s).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\Phi(\xi) \to \frac{1}{(s - 1)} = l
For s greater than 1, the integral tends to the limit 1/(s-1), so the series of n^(-s) converges with sum at most s/(s-1).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x}
For s equal to 1, Phi(xi) is the integral of 1/x from 1 to xi, which tends to infinity, so the harmonic series diverges.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\ds\Phi(\xi) > n\int_{1}^{2} \frac{du}{u}
For xi greater than 2^n, the integral of 1/x from 1 to xi exceeds n times the integral of 1/u from 1 to 2, which shows it tends to infinity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{2^{r}}^{2^{r+1}} \frac{dx}{x} = \int_{1}^{2} \frac{du}{u}
Substituting x = 2^r u shows that the integral of 1/x over each interval from 2^r to 2^(r+1) equals the same value as over 1 to 2.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim_{x \to \infty} \int_{1}^{x} \phi(t)\, dt = l
If the integral of a positive decreasing function from 1 up to x approaches a limit l as x grows without bound, the infinite integral is convergent with value l.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim_{x \to \infty} \int_{a}^{x} \phi(t)\, dt = l
The infinite integral from a to infinity of phi is defined as convergent with value l when the integral from a to x tends to l as x tends to infinity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{x}\phi(t)\, dt = \Phi(x)
Writing Phi(x) for the integral of phi from a to x defines the integral function, and the infinite integral converges, diverges or oscillates according as Phi(x) tends to a limit, to infinity, or oscillates.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\infty} \phi(x)\, dx = \int_{a}^{b} \phi(x)\, dx + \int_{b}^{\infty}\phi(x)\, dx
If the infinite integral from a converges and b > a, the infinite integral splits into a finite integral from a to b plus the infinite integral from b.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{x} \phi(t)\, dt < K
The infinite integral from a converges exactly when the integral from a to x stays below some fixed constant K for all x greater than a.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\infty} \psi(x)\, dx \leq K\int_{a}^{\infty} \phi(x)\, dx
If psi(x) is at most K times phi(x) beyond a and the integral of phi converges, then the integral of psi converges and is at most K times the integral of phi.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim x^{s}\phi(x) = l
If x^s times phi(x) tends to a positive limit l, the infinite integral of phi is convergent when s > 1 and divergent when s <= 1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\xi} \phi(x)\, dx < \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}}
For the constructed function of the remark, the integral up to any xi is bounded by the convergent sum of 1/(n+1)^2, so the infinite integral converges.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\xi} \phi(x)\, dx = \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt
Under the substitution x = f(t), with a = f(b) and xi = f(tau), the integral over x equals the integral over t of phi{f(t)} f'(t).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\infty} \phi(x)\, dx = \int_{b}^{c} \phi\{f(t)\}f'(t)\, dt
The infinite integral in x equals the integral in t from b to c, which is an infinite integral of the second kind when the integrand is infinite at t = c.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\xi} f(x)\phi'(x)\, dx = f(\xi)\phi(\xi) - f(a)\phi(a) - \int_{a}^{\xi} f'(x)\phi(x)\, dx
Integration by parts: the integral of f times the derivative of phi equals the boundary terms f phi evaluated at xi and a, minus the integral of f' times phi.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{A} (x - a)^{-s}\, dx = \lim_{\epsilon\to +0} \int_{a+\epsilon}^{A} (x - a)^{-s}\, dx
An integral whose integrand tends to infinity at the lower limit a is defined as the limit, as epsilon tends to zero from above, of the integral starting at a + epsilon.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{1/A}^{\eta} y^{s-2}\, dy = \int_{1/\eta}^{A} x^{-s}\, dx
Under the substitution y = 1/x, the integral of y to the power s-2 from 1/A to eta equals the integral of x to the power -s from 1/eta to A.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{a}^{\infty} \phi(x)\, dx = \lim_{\tau\to c} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt
Equation (4) of the substitution rule: the infinite integral equals the limit of the transformed integral as tau approaches c.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
J = \int_{1}^{7} (x^{2} - 6x + 13)\, dx
J is defined as the definite integral of x squared minus six x plus thirteen from 1 to 7.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
J = 48
Direct integration gives the value of the integral J as 48.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
y = x^{2} - 6x + 13
The substitution sets y equal to x squared minus six x plus thirteen.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
x = 3 ± \sqrtp{y - 4}
Solving the substitution for x gives two branches, x equal to 3 plus or minus the square root of y minus 4.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
J = \int_{1}^{7} y\, dx = \int_{8}^{4} \left\{-\frac{y}{2\sqrtp{y - 4}}\right\} dy + \int_{4}^{20} \frac{y}{2\sqrtp{y - 4}}\, dy
The correct value of J is the sum of two integrals in y, one on each branch of the substitution, with the sign of dx/dy chosen on each branch.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\pi} dx = \pi
The integral of 1 from 0 to pi equals pi.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
x = \arcsin y
The substitution x equals the inverse sine of y.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
dx/dy = 1/\sqrtp{1 - y^{2}}
When x is between 0 and a half pi, dx/dy equals one over the square root of one minus y squared.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
dx/dy = -1/\sqrtp{1 - y^{2}}
When x is between a half pi and pi, dx/dy equals minus one over the square root of one minus y squared.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|u_{n}| = \alpha_{n}
alpha_n is defined as the modulus of u_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = v_{n} - w_{n}
Each term u_n is the positive part v_n minus the negative part w_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\alpha_{n} = v_{n} + w_{n}
The modulus alpha_n is the sum of the positive part and the negative part of u_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\tsum u'_{n} = \tsum v'_{n} - \tsum w'_{n} = \tsum v_{n} - \tsum w_{n} = \tsum u_{n}
A rearrangement of an absolutely convergent series has the same sum as the original series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum_{0}^{N} u_{n} = \sum_{0}^{N} v_{n} - \sum_{0}^{N} w_{n}
The partial sum of u_n up to N equals the partial sum of v_n minus the partial sum of w_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|v_{0}| + |v_{1}| + \dots + |v_{n}| < |u_{0}| + \dots + |u_{n}|
The sum of the moduli of the v terms is less than the sum of the moduli of the u terms.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{2n+1} - s_{2n-1} = \phi_{2n} - \phi_{2n+1}\geq 0
The odd partial sums of an alternating series with steadily decreasing terms increase at each step.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{2n} - s_{2n-2} = -(\phi_{2n-1} - \phi_{2n}) \leq 0
The even partial sums of an alternating series with steadily decreasing terms decrease at each step.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{n} = \phi_{0} - \phi_{1} + \phi_{2} - \dots + (-1)^{n}\phi_{n}
s_n is defined as the nth partial sum of the alternating series of phi values.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim (s_{2n+1} - s_{2n}) = \lim (-1)^{2n+1} \phi_{2n+1} = 0
The difference between successive partial sums of an alternating series tends to zero.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{1} = \phi_{0} - \phi_{1}
The first partial sum of the alternating series is phi_0 minus phi_1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
a_{0}\phi_{0} + a_{1}\phi_{1} + \dots + a_{n}\phi_{n} = s_{0}(\phi_{0} - \phi_{1}) + s_{1}(\phi_{1} - \phi_{2}) + \dots + s_{n-1}(\phi_{n-1} - \phi_{n}) + s_{n}\phi_{n}
The partial sum of the products a_k phi_k can be rewritten using partial sums s_k of the a's and differences of the phi's.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{n} = a_{0} + a_{1} + \dots + a_{n}
s_n is the nth partial sum of the series of a's.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|s_{\nu}| < K
The partial sums s_nu of a series that converges or oscillates finitely are bounded by a constant K.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
s_{m, \nu} = a_{m} + a_{m+1} + \dots + a_{\nu}
s_{m,nu} is the partial sum of the a's from index m to index nu.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|a_{m}\phi_{m} + a_{m+1}\phi_{m+1} + \dots + a_{n}\phi_{n}| < \DELTA \phi_{m} \leq \DELTA \phi_{1}
The modulus of a block of terms a_k phi_k is less than a positive number Delta times phi_1, when m is large enough.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
z = \Cis\theta
z is defined as cos theta plus i sin theta, so that the modulus of z is 1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|s_{n} + it_{n}| = \left|\frac{1 - z^{n}}{1 - z}\right| \leq \frac{1 + |z^{n}|}{|1 - z|} \leq \frac{2}{|1 - z|}
The partial sums of cos n theta and sin n theta are bounded by 2 over the modulus of 1 minus z, when z is not 1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\tsum u_{n} = \tsum (v_{n} + iw_{n})
A complex series is written as the sum of its real parts v_n plus i times its imaginary parts w_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = v_{n} + iw_{n}
The nth term of a complex series is its real part plus i times its imaginary part.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|u_{n}| = \sqrtp{v_{n}^{2} + w_{n}^{2}}
The modulus of a complex term is the square root of the sum of the squares of its real and imaginary parts.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|u_{n}| = \sqrtp{v_{n}^{2} + w_{n}^{2}} \leq |v_{n}| + |w_{n}|
The modulus of a complex term is at most the sum of the moduli of its real and imaginary parts.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|v_{n}| \leq \sqrtp{v_{n}^{2} + w_{n}^{2}}
The modulus of the real part is at most the modulus of the complex term.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|u_{n+1}|/|u_{n}| = |z|/(n + 1) \to 0
For the series of z^n over n factorial, the ratio of successive term moduli tends to zero for every z, giving convergence for all z.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|u_{n+1}|/|u_{n}| = (n + 1)|z|
For the series of n! z^n, the ratio of successive term moduli grows without limit unless z is zero.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim \left[\frac{1}{2n + 1} - \frac{1}{2n + 2} + \frac{1}{2n + 3} - \dots + \frac{1}{4n - 1} - \frac{1}{4n}\right] = 0
The alternating bracket of reciprocals tends to zero as n tends to infinity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim \left(\frac{1}{2n + 2} + \frac{1}{2n + 4} + \dots + \frac{1}{4n}\right) = \tfrac{1}{2} \lim \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + (r/n)} = \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x}
The limit of the sum of reciprocals from 2n+2 to 4n equals one half the integral of 1/x from 1 to 2.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim t_{3n} = s + \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x}
The sum of the rearranged series is s plus one half the integral of 1/x from 1 to 2, not s.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim a_{n}z_{1}^{n} = 0
If the power series converges at z_1, its terms a_n z_1^n tend to zero.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
|a_{n}z^{n}| = |a_{n}z_{1}^{n}| \left(\frac{r}{r_{1}}\right)^{n} < K \left(\frac{r}{r_{1}}\right)^{n}
For |z| = r < r_1 the terms of the power series are bounded by a geometric term, so comparison with a convergent geometrical series gives absolute convergence.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
z = \cos\theta + i\sin\theta
A complex number of modulus one written in terms of its angle θ, cos θ + i sin θ.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\lim |a_{n+1}z^{n+1}|/|a_{n}z^{n}| = \lambda |z|
The ratio of successive absolute terms of a power series tends to λ|z|, so by the ratio test the series converges when |z| < 1/λ.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\frac{|a_{n+1}|}{|a_{n}|} = \frac{|m - n|}{n + 1} \to 1
For the binomial series the ratio of successive absolute coefficients tends to 1, so its radius of convergence is unity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
1/(1 - z)^{2} = 1 + 2z + 3z^{2} + \dots
For |z| < 1 the square of the reciprocal of 1 - z expands as a power series with coefficients 1, 2, 3, ...
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\frac{1}{(1 - z)^{m}} = 1 + mz + \frac{m(m + 1)}{1·2} z^{2} + \dots
For |z| < 1 and positive integer m, the reciprocal power 1/(1 - z)^m expands in an infinite power series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(m, z) = 1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots
Defines the binomial series f(m, z) as a power series in z with binomial coefficients.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(m, z)f(m', z) = f(m + m', z)
Multiplying two binomial series with exponents m and m' gives the binomial series with exponent m + m' for |z| < 1; this is the basis of Euler's proof of the binomial theorem.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(z)f(z') = f(z + z')
The series f(z) = 1 + z + z^2/2! + ... satisfies the addition law f(z)f(z') = f(z + z') for all z, z'.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
C(z) = 1 - \frac{z^{2}}{2!} + \frac{z^{4}}{4!} - \dots
Defines C(z) as the power series whose even-power pattern gives the cosine series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
S(z) = z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} - \dots
Defines S(z) as the power series whose odd-power pattern gives the sine series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
C(z + z') = C(z)C(z') - S(z)S(z')
The series C satisfies the addition formula for the cosine.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
S(z + z') = S(z)C(z') + C(z)S(z')
The series S satisfies the addition formula for the sine.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\{C(z)\}^{2} + \{S(z)\}^{2} = 1
The squares of the cosine-type and sine-type series sum to one.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = v_{n} = \frac{(-1)^{n}}{\sqrtp{n + 1}}
The two series considered have terms (-1)^n over the square root of n+1; they are conditionally, not absolutely, convergent.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
w_{n} = (-1)^{n} \sum_{r=0}^{n} \frac{1}{\sqrtb{(r + 1)(n + 1 - r)}}
The Cauchy product coefficient of the two series of example 9, which does not tend to zero and so the product series fails to converge.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum u_{n} × \sum v_{n} = \sum w_{n}
The product of two absolutely convergent series equals the series of Cauchy-product terms w_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
c_{n} = a_{0}b_{n} + a_{1}b_{n-1} + \dots + a_{n}b_{0}
The coefficient of z^n in the product of two power series is this sum of products of coefficients.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(z)/(1 - z) = \sum s_{n}z^{n}
Dividing a convergent power series by 1 - z gives the power series whose coefficients are the partial sums of the a_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum_{1}^{\infty} \frac{n^{2} + 9n + 5}{(n + 1)(2n + 3)(2n + 5)(n + 4)} = \frac{5}{36}
The infinite series of this rational general term has sum 5/36.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = \frac{x^{n} - x^{-n-1}}{(x^{n} + x^{-n})(x^{n+1} + x^{-n-1}) }
The general term of the series of example 13.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\frac{x^{n} - x^{-n-1}}{(x^{n} + x^{-n})(x^{n+1} + x^{-n-1}) } = \frac{1}{x - 1} \left(\frac{1}{x^{n} + x^{-n}} - \frac{1}{x^{n+1} + x^{-n-1}}\right)
The general term splits as a difference of two terms, which makes the series telescope.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
a_{n} + p_{1}a_{n-1} + p_{2}a_{n-2} + \dots + p_{k}a_{n-k} = 0
A recurring series has coefficients satisfying this linear relation with constant coefficients p_1,...,p_k.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
(1 + p_{1}z + p_{2}z^{2} + \dots + p_{k}z^{k})f(z) = P_{0} + P_{1}z + \dots + P_{k-1}z^{k-1}
Multiplying a recurring series by its scale of relation gives a polynomial of degree k-1, so the series is a rational function.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
a_{n} - a_{n-1} - 8a_{n-2} + 12a_{n-3} = 0
Example of a linear difference equation with constant coefficients.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
a_{n} = 2^{n}\{A_{1} + (n + 1) A_{2}\} + (-3)^{n} B
The general solution of the example difference equation, with constants fixed by a_0, a_1, a_2.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} - 2\cos\theta u_{n-1} + u_{n-2} = 0
The difference equation whose solutions are sinusoidal in n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
u_{n} = A\cos n\theta + B\sin n\theta
The general solution of the difference equation with cos θ, with arbitrary constants A and B.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(n) + f(n - 1) + f(n - 2) = 0
The coefficients of z/(1 + z + z^2) satisfy this three-term relation.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
f(n) = (\omega_{3}^{n} - \omega_{3}^{2n})/(\omega_{3} - \omega_{3}^{2})
Closed form for the coefficient f(n), using a complex cube root of unity ω_3.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
z/(1 + z + z^{2}) = z(1 - z)/(1 - z^{3})
The rational function equals z(1 - z)/(1 - z^3), used to verify the coefficients of f(n).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
p_{n} = \frac{1}{2} (p_{n-1} + p_{n-2})
The probability of exactly reaching total n satisfies this recurrence.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\frac{1}{3}\{2 + (-\frac{1}{2})^{n}\}
The chance that the player's score makes exactly the total n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\frac{1}{a + 1} + \frac{1}{a + 2} + \dots + \frac{1}{a + n} = \binom{n}{1}\frac{1}{a + 1} - \binom{n}{2}\frac{1!}{(a + 1)(a + 2)} + \dots
Identity expressing a sum of reciprocals as a sum of binomial-weighted partial-fraction terms, for positive integer n and a not in {-1,...,-n}.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{1} x^{a}\frac{1 - x^{n}}{1 - x}\, dx = \int_{0}^{1} (1 - x)^{a}\{1 - (1 - x)^{n}\}\frac{dx}{x}
Two integral forms that give the same value, used to prove the preceding identity when a > -1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\sum_{0}^{\infty} \frac{z^{n}}{n!} \sum_{1}^{\infty} \frac{(-1)^{n-1}z^{n}}{n·n!} = \sum_{1}^{\infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) \frac{z^{n}}{n!}
The product of the exponential-type series with the logarithm-type series equals a series with harmonic-number coefficients.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
(A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1})/n \to AB
If A_n tends to A and B_n tends to B, the Cesàro-type mean of the convolution tends to AB.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
c_{n} = a_{1}b_{n} + a_{2}b_{n-1} + \dots + a_{n}b_{1}
Definition of the convolution terms c_n of two series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
C_{n} = a_{1}B_{n} + a_{2}B_{n-1} + \dots + a_{n}B_{1}
The n-th partial sum of the convolution equals this sum of terms with partial sums of the b_n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
C_{1} + C_{2} + \dots + C_{n} = A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1}
The sum of the first n partial sums of the convolution equals the convolution of partial sums of the two series.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
(C_{1} + C_{2} + \dots + C_{n})/n \to AB
The Cesàro mean of the partial sums of the product series tends to AB; hence if the product series converges its sum is AB.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{-1}^{1} \frac{dx}{(a - x) \sqrtp{1 - x^{2}}} = \frac{\pi}{\sqrtp{a^{2} - 1}}
Evaluates a definite integral with a convergent integrand for a > 1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\infty} \frac{dx}{\{\sqrtp{x^{2} + 1} + x\}^{n}} = \int_{0}^{\infty} \{\sqrtp{x^{2} + 1} - x\}^{n}\, dx = \frac{n}{n^{2} - 1}
The two infinite integrals are equal and have value n/(n^2 - 1) for n > 1.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
2y = ax - (b/x)
Substitution relation between y and x, with a, b positive, which increases steadily from -infinity to infinity.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
2y = ax + (b/x)
Substitution relation between y and x, where two values of x correspond to each y greater than sqrt(ab).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\pi} f(\sec\tfrac{1}{2}x + \tan\tfrac{1}{2}x)\frac{dx}{\sqrtp{\sin x}} = \int_{0}^{\pi} f(\cosec x)\frac{dx}{\sqrtp{\sin x}}
Transformation formula equating two integrals with the same weight 1/sqrt(sin x).
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\infty} \frac{dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \frac{\pi}{2ab(a + b)}
Evaluates an infinite integral for positive a and b.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} + a^{2})(x^{2} + b^{2})} = \frac{\pi}{2(a + b)}
Evaluates an infinite integral for positive a and b.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
A = \beta + \sqrtp{\alpha\gamma}
Defines the constant A used in the closed forms of the quartic-denominator integrals.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} - a^{2})^{2} + b^{2}x^{2}} = \frac{\pi}{2b}
Evaluates an infinite integral for positive b.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{0}^{\infty} \phi(x)\, dx = \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}}
The integral of the function phi from the end of section 178 equals the sum of 1/(n+1)^2 over n.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x - y}{(x + y)^{3}}\, dy\right) = -1
Iterated infinite integral with the outer variable x; its order-reversed value is 1, showing the order of integration matters.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x - y}{(x + y)^{3}}\, dx\right) = 1
Iterated infinite integral with the outer variable y; its value differs in sign from the reversed order.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\, dy\right) = -\tfrac{1}{4}\pi
Iterated infinite integral with outer variable x; its reversed order gives +pi/4.
THE CONVERGENCE OF INFINITE SERIES AND \\ INFINITE INTEGRALS
\int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\, dx\right) = \tfrac{1}{4}\pi
Iterated infinite integral with outer variable y; the two orders give different values.

Problems

Exercise LXXVII

Exercise LXXVII, problem 1, p. 337
Employ the ‘general principle of convergence’ ([§]84) to prove the theorem that an absolutely convergent series is convergent. [Since $\sum |u_{n}|$ is convergent, we can, when any positive number $\DELTA$ is assigned, choose $n_{0}$ so that |u_n_1+1| + |u_n_1+2| + …+ |u_n_2| < when $n_{2} > n_{1} \geq n_{0}$. *A fortiori* |u_n_1+1 + u_n_1+2 + …+ u_n_2| < , and therefore $\sum u_{n}$ is convergent.]
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Exercise LXXVII, problem 2, p. 337
If $\sum a_{n}$ is a convergent series of positive terms, and $|b_{n}|\leq Ka_{n}$, then $\sum b_{n}$ is absolutely convergent.
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Exercise LXXVII, problem 3, p. 337
If $\sum a_{n}$ is a convergent series of positive terms, then the series $\sum a_{n}x^{n}$ is absolutely convergent when $-1 \leq x \leq 1$.
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Exercise LXXVII, problem 4, p. 337
If $\sum a_{n}$ is a convergent series of positive terms, then the series $\sum a_{n} \cos n\theta$, $\sum a_{n}\sin n\theta$ are absolutely convergent for all values of $\theta$. [Examples are afforded by the series $\sum r^{n}\cos n\theta$, $\sum r^{n}\sin n\theta$ of [§]88.]
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Exercise LXXVII, problem 5, p. 337
Any series selected from the terms of an absolutely convergent series is absolutely convergent. [For the series of the moduli of its terms is a selection from the series of the moduli of the terms of the original series.]
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Exercise LXXVII, problem 6, p. 337
Prove that if $\sum |u_{n}|$ is convergent then |u_n| |u_n|, and that the only case to which the sign of equality can apply is that in which every term has the same sign.
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Exercise LXXVIII

Exercise LXXVIII, problem 1, p. 340
The series gather* 1 - 12 + 13 - 14 + …,0pt minus 3pt1 - 12 + 13 - 14 + …, (-1)^n(n + a),0pt minus 3pt(-1)^nn + a,0pt minus 3pt(-1)^n(n + a),0pt minus 3pt(-1)^n(n + a)^2, gather* where $a > 0$, are conditionally convergent.
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Exercise LXXVIII, problem 2, p. 340
The series $\sum(-1)^{n}(n + a)^{-s}$, where $a > 0$, is absolutely convergent if $s > 1$, conditionally convergent if $0 < s \leq 1$, and oscillatory if $s \leq 0$.
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Exercise LXXVIII, problem 3, p. 340
The sum of the series of [§]188 lies between $s_{n}$ and $s_{n+1}$ for all values of $n$; and the error committed by taking the sum of the first $n$ terms instead of the sum of the whole series is numerically not greater than the modulus of the $(n + 1)$th term.
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Exercise LXXVIII, problem 4, p. 340
Consider the series (-1)^nn + (-1)^n, which we suppose to begin with the term for which $n = 2$, to avoid any difficulty as to the definitions of the first few terms. This series may be written in the form [ (-1)^nn + (-1)^n - (-1)^nn + (-1)^nn] or (-1)^nn - 1n + (-1)^nn = (_n - _n), say. The series $\sum \psi_{n}$ is convergent; but $\sum \chi_{n}$ is divergent, as all its terms are positive, and $\lim n\chi_{n} = 1$. Hence the original series is divergent, although it is of the form $\phi_{2} - \phi_{3} + \phi_{4} - \dots$, where $\phi_{n} \to 0$. This example shows that the condition that $\phi_{n}$ should tend *steadily* to zero is essential to the truth of the theorem. The reader will easily verify that $\sqrtp{2n + 1} - 1 < \sqrtp{2n} + 1$, so that this condition is not satisfied.
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Exercise LXXVIII, problem 5, p. 340
If the conditions of [§]188 are satisfied except that $\phi_{n}$ tends steadily to a positive limit $l$, then the series $\sum (-1)^{n}\phi_{n}$ oscillates finitely.
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Exercise LXXVIII, problem 6, p. 340
**of the sum of a conditionally convergent series by rearrangement of the terms.** Let $s$ be the sum of the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$, and $s_{2n}$ the sum of its first $2n$ terms, so that $\lim s_{2n} = s$. Now consider the series 1 + 13 - 12 + 15 + 17 - 14 + …(1) in which two positive terms are followed by one negative term, and let $t_{3n}$ denote the sum of the first $3n$ terms. Then align* t_3n &= 1 + 13 + …+ 14n-1 - 12 - 14 - …- 12n &= s_2n + 12n + 1 + 12n + 3 + …+ 14n - 1. align* Now [12n + 1 - 12n + 2 + 12n + 3 - … + 14n - 1 - 14n] = 0, 0.375em plus 0.75em minus 0.25emsince the sum of the terms inside the bracket is clearly less than $n/(2n + 1)(2n + 2)$; and (12n + 2 + 12n + 4 + …+ 14n) = 12 1n _r=1^n 11 + (r/n) = 12 _1^2 dxx, by [§§]156 and 158. Hence t_3n = s + 12 _1^2 dxx, [pg]342 and it follows that the sum of the series (1) is not $s$, but the right-hand side of the last equation. Later on we shall give the actual values of the sums of the two series: see [§]213 and Ch.IX, [misc:IX]Misc. Ex. 19. It can indeed be proved that a conditionally convergent series can always be so rearranged as to converge to any sum whatever, or to diverge to $\infty$ or to $-\infty$. For a proof we may refer to Bromwich’s *Infinite Series*, p. 68.
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Exercise LXXVIII, problem 7, p. 340
The series 1 + 13 - 12 + 15 + 17 - 14 + … diverges to $\infty$. [Here t_3n = s_2n + 12n + 1 + 12n + 3 + … + 14n - 1 > s_2n + n4n - 1, where $s_{2n} = 1 - \dfrac{1}{\sqrt{2}} + \dots - \dfrac{1}{\DPtypo{\sqrt{2n}}{\sqrtp{2n}}}$, which tends to a limit as $n \to \infty$.]
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Exercise LXXIX

Exercise LXXIX, problem 1, p. 343
Dirichlet’s and Abel’s Tests may also be established by means of the general principle of convergence ([§]84). Let us suppose, for example, that the conditions of Abel’s Test are satisfied. We have identically 0pt multline* a_m_m + a_m+1_m+1 + …+ a_n_n = s_m, m(_m - _m+1) + s_m, m+1(_m+1 - _m+2) + …+ s_m, n-1(_n-1 - _n) + s_m, n_n…, (1) multline*% where s_m, = a_m + a_m+1 + …+ a_. The left-hand side of (1) therefore lies between $h\phi_{m}$ and $H\phi_{m}$, where $h$ and $H$ are the algebraically least and greatest of $s_{m, m}$, $s_{m, m+1}$, …, $s_{m, n}$. But, given any positive number $\DELTA$, we can choose $m_{0}$ so that $|s_{m, \nu}| < \DELTA$ when $m \geq m_{0}$, and so |a_m_m + a_m+1_m+1 + …+ a_n_n| < _m _1 when $n > m \geq m_{0}$. Thus the series $\sum a_{n}\phi_{n}$ is convergent.
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Exercise LXXIX, problem 2, p. 343
The series $\sum \cos n\theta$ and $\sum \sin n\theta$ oscillate finitely when $\theta$ is not a multiple of $\pi$. For, if we denote the sums of the first $n$ terms of the two series by $s_{n}$ and $t_{n}$, and write $z = \Cis\theta$, so that $|z| = 1$ and $z \neq 1$, we have |s_n + it_n| = |1 - z^n1 - z| 1 + |z^n||1 - z| 2|1 - z|; and so $|s_{n}|$ and $|t_{n}|$ are also not greater than $2/|1 - z|$. That the series are not actually convergent follows from the fact that their $n$th terms do not tend to zero (xxiv. 7, 8). The sine series converges to zero if $\theta$ is a multiple of $\pi$. The cosine series oscillates finitely if $\theta$ is an odd multiple of $\pi$ and diverges if $\theta$ is an even multiple of $\pi$. It follows that *if $\theta_{n}$ is a positive function of $n$ which tends steadily to zero as $n \to \infty$, then the series _n n,0pt minus 3pt_n n are convergent*, except perhaps the first series when $\theta$ is a multiple of $2\pi$. In this case the first series reduces to $\sum \phi_{n}$, which may or may not be convergent: the second series vanishes identically. If $\sum \phi_{n}$ is convergent then both series are absolutely convergent (% [examples:lxxvii]Ex. lxxvii%. 4) for all values of $\theta$, and the whole interest of the result lies in its application to the case in which $\sum \phi_{n}$ is divergent. And in this case the series above written are conditionally and *not* absolutely convergent, as will be proved in % [examples:lxxix]Ex. lxxix%. 6. If we put $\theta = \pi$ in the cosine series we are led back to the result of [§]188, since $\cos n\pi = (-1)^{n}$.
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Exercise LXXIX, problem 3, p. 343
The series $\sum n^{-s} \cos n\theta$, $\sum n^{-s} \sin n\theta$ are convergent if $s > 0$, unless (in the case of the first series) $\theta$ is a multiple of $2\pi$ and $0 < s \leq 1$.
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Exercise LXXIX, problem 4, p. 343
The series of Ex. 3 are in general absolutely convergent if $s > 1$, conditionally convergent if $0 < s \leq 1$, and oscillatory if $s \leq 0$ (finitely if $s = 0$ and infinitely if $s < 0$). Mention any exceptional cases.
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Exercise LXXIX, problem 5, p. 343
If $\sum a_{n}n^{-s}$ is convergent or oscillates finitely, then $\sum a_{n}n^{-t}$ is convergent when $t > s$.
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Exercise LXXIX, problem 6, p. 343
If $\phi_{n}$ is a positive function of $n$ which tends steadily to $0$ as $n \to \infty$, and $\sum \phi_{n}$ is divergent, then the series $\sum \phi_{n} \cos n\theta$, $\sum \phi_{n} \sin n\theta$ are *not* absolutely convergent, except the sine-series when $\theta$ is a multiple of $\pi$. [For suppose, *e.g.*, that $\sum \phi_{n} |\cos n\theta|$ is convergent. Since $\cos^{2} n\theta \leq |\cos n\theta|$, it follows that $\sum \phi_{n} \cos^{2} n\theta$ or 12 _n (1 + 2n) is convergent. But this is impossible, since $\sum \phi_{n}$ is divergent and $\sum \phi_{n} \cos 2n\theta$, by Dirichlet’s Test, convergent, unless $\theta$ is a multiple of $\pi$. And in this case it is obvious that $\sum \phi_{n} |\cos n\theta|$ is divergent. The reader should write out the corresponding argument for the sine-series, noting where it fails when $\theta$ is a multiple of $\pi$.]
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Exercise LXXX

Exercise LXXX, problem 1, p. 347
The series $1 + az + a^{2}z^{2} + \dots$, where $a > 0$, has a radius of convergence equal to $1/a$. It does not converge anywhere on its circle of convergence, diverging when $z = 1/a$ and oscillating finitely at all other points on the circle.
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Exercise LXXX, problem 2, p. 347
The series $\dfrac{z}{1^{2}} + \dfrac{z^{2}}{2^{2}} + \dfrac{z^{3}}{3^{2}} + \dots$ has its radius of convergence equal to $1$; it converges absolutely at all points on its circle of convergence.
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Exercise LXXX, problem 3, p. 347
More generally, if $|a_{n+1}|/|a_{n}| \to \lambda$, or $|a_{n}|^{1/n} \to \lambda$, as $n \to \infty$, then the series $a_{0} + a_{1}z + a_{2}z^{2} + \dots$ has $1/\lambda$ as its radius of convergence. In the first case |a_n+1z^n+1|/|a_nz^n| = |z|, which is less or greater than unity according as $|z|$ is less or greater than $1/\lambda$, so that we can use D’Alembert’sd’Alembert’s Test ([§]168, 3). In the second case we can use Cauchy’s Test ([§]168, 2) similarly.
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Exercise LXXX, problem 4, p. 347
**logarithmic series.** The series z - 12 z^2 + 13 z^3 - … is called (for reasons which will appear later) the ‘logarithmic’ series. It follows from Ex. 3 that its radius of convergence is unity. When $z$ is on the circle of convergence we may write $z = \cos\theta + i\sin\theta$, and the series assumes the form - 12 2+ 13 3- …+ i(- 12 2+ 13 3- …). The real and imaginary parts are both convergent, though not absolutely convergent, unless $\theta$ is an odd multiple of $\pi$ (lxxix. 3, 4). If $\theta$ is an odd multiple of $\pi$ then $z = -1$, and the series assumes the form $-1 - \frac{1}{2} - \frac{1}{3} - \dots$, and so diverges to $-\infty$. Thus the logarithmic series converges at all points of its circle of convergence except the point $z = -1$.
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Exercise LXXX, problem 5, p. 347
**binomial series.** Consider the series 1 + mz + m(m - 1)2! z^2 + m(m - 1)(m - 2)3! z^3 + … If $m$ is a positive integer then the series terminates. In general |a_n+1||a_n| = |m - n|n + 1 1, so that the radius of convergence is unity. We shall not discuss here the question of its convergence on the circle, which is a little more difficult. See Bromwich, *Infinite Series*, pp. 225 *et seq.*; Hobson, *Plane Trigonometry* (3rd edition), pp. 268 *et seq.*
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Exercise LXXXI

Exercise LXXXI, problem 1, p. 349
If $|z|$ is less than the radius of convergence of either of the series $\sum a_{n}z^{n}$, $\sum b_{n}z^{n}$, then the product of the two series is $\sum c_{n}z^{n}$, where $c_{n} = a_{0}b_{n} + a_{1}b_{n-1} + \dots + a_{n}b_{0}$.
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Exercise LXXXI, problem 2, p. 349
0.375em plus 0.75em minus 0.25emIf the radius of convergence of $\sum a_{n}z^{n}$ is $R$, and $f(z)$ is the sum of the series when $|z| < R$, and $|z|$ is less than either $R$ or unity, then $f(z)/(1 - z) = \sum s_{n}z^{n}$, where $s_{n} = a_{0} + a_{1} + \dots + a_{n}$.
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Exercise LXXXI, problem 3, p. 349
Prove, by squaring the series for $1/(1 - z)$, that $1/(1 - z)^{2} = 1 + 2z + 3z^{2} + \dots$ if $|z| < 1$.
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Exercise LXXXI, problem 4, p. 349
Prove similarly that $1/(1 - z)^{3} = 1 + 3z + 6z^{2} + \dots$, the general term being $\frac{1}{2}(n + 1)(n + 2)z^{n}$.
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Exercise LXXXI, problem 5, p. 349
**Binomial Theorem for a negative integral exponent.** If $|z| < 1$, and $m$ is a positive integer, then 1(1 - z)^m = 1 + mz + m(m + 1)1·2 z^2 + … + m(m + 1) …(m + n - 1)1·2 …n z^n + …. [Assume the truth of the theorem for all indices up to $m$. Then, by Ex. 2, $1/(1 - z)^{m+1} = \sum s_{n}z^{n}$, where align* %[** TN: Set on a single line in the original] s_n &= 1 + m + m(m + 1)1·2 + … + m(m + 1) …(m + n - 1)1·2 …n &= (m + 1)(m + 2) …(m + n)1·2 …n, align* as is easily proved by induction.]
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Exercise LXXXI, problem 6, p. 349
Prove by multiplication of series that if f(m, z) = 1 + m1 z + m2 z^2 + …, and $|z| < 1$, then $f(m, z)f(m', z) = f(m + m', z)$. [This equation forms the basis of Euler’s proof of the Binomial Theorem. The coefficient of $z^{n}$ in the product series is m’n + m1 m’n - 1 + m2 m’n - 2 + … + mn - 1 m’1 + mn. [pg]350 This is a polynomial in $m$ and $m'$: but when $m$ and $m'$ are positive integers this polynomial must reduce to $\dbinom{m + m'}{k}$ in virtue of the Binomial Theorem for a positive integral exponent, and if two such polynomials are equal for all positive integral values of $m$ and $m'$ then they must be equal identically.]
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Exercise LXXXI, problem 7, p. 349
If $f(z) = 1 + z + \dfrac{z^{2}}{2!} + \dots$ then $f(z)f(z') = f(z + z')$. [For the series for $f(z)$ is absolutely convergent for all values of $z$: and it is easy to see that if $u_{n} = \dfrac{z^{n}}{n!}$, $v_{n} = \dfrac{z'^{n}}{n!}$, then $w_{n} = \dfrac{(z + z')^{n}}{n!}$.]
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Exercise LXXXI, problem 8, p. 349
If C(z) = 1 - z^22! + z^44! - …,0pt minus 3ptS(z) = z - z^33! + z^55! - …, then C(z + z’) = C(z)C(z’) - S(z)S(z’),0pt minus 3ptS(z + z’) = S(z)C(z’) + C(z)S(z’), and C(z)^2 + S(z)^2 = 1.
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Exercise LXXXI, problem 9, p. 349
**of the Multiplication Theorem.** That the theorem is not always true when $\sum u_{n}$ and $\sum v_{n}$ are not *absolutely* convergent may be seen by considering the case in which u_n = v_n = (-1)^nn + 1. Then w_n = (-1)^n _r=0^n 1(r + 1)(n + 1 - r). But $\sqrtb{(r + 1)(n + 1 - r)} \leq \frac{1}{2}(n + 2)$, and so $|w_{n}| > (2n + 2)/(n + 2)$, which tends to $2$; so that $\sum w_{n}$ is certainly not convergent.
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Exercise LXVII

Exercise LXVII, problem 1, p. 311
Apply Cauchy’s and d’Alembert’s tests (as311 specialised in 4 above) to the series $\sum n^{k} r^{n}$, where $k$ is a positive rational number.
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Exercise LXVII, problem 10, p. 311
The series $1 + r + \dfrac{r^{2}}{2!} + \dfrac{r^{3}}{3!} + \dots$ and $1 + r + \dfrac{r^{2}}{2^{2}} + \dfrac{r^{3}}{3^{3}} + \dots$ are convergent for all positive values of $r$.
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Exercise LXVII, problem 11, p. 311
If $\sum u_{n}$ is convergent then so are $\sum u_{n}^{2}$ and $\sum u_{n}/(1 + u_{n})$.
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Exercise LXVII, problem 12, p. 311
If $\sum u_{n}^{2}$ is convergent then so is $\sum u_{n}/n$.
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Exercise LXVII, problem 13, p. 311
Show that %[** TN: In-line in the original] 1 + 13^2 + 15^2 + … = 34(1 + 12^2 + 13^2 + …) and 1 + 12^2 + 13^2 + 15^2 + 16^2 + 17^2 + 19^2 + … = 1516 (1 + 12^2 + 13^2 + …).
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Exercise LXVII, problem 14, p. 311
Prove by a *reductio ad absurdum* that $\sum (1/n)$ is divergent.
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Exercise LXVII, problem 2, p. 311
Consider the series $\sum(An^{k} + Bn^{k-1} + \dots + K) r^{n}$.
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Exercise LXVII, problem 3, p. 311
Consider An^k + Bn^k-1 + …+ K n^l + n^l-1 + …+ r^n0pt minus 3pt(A > 0, > 0).
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Exercise LXVII, problem 4, p. 311
We have seen (Ch.IV, [misc:IV]Misc. Ex. 17) that the series 1n(n + 1),0pt minus 3pt1n(n + 1)…(n + p) are convergent. Show that Cauchy’s and d’Alembert’s tests both fail when applied to them.
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Exercise LXVII, problem 5, p. 311
Show that the series $\sum n^{-p}$, where $p$ is an integer not less than $2$, is convergent.
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Exercise LXVII, problem 6, p. 311
Show that the series An^k + Bn^k-1 + …+ K n^l + n^l-1 + …+ is convergent if $l > k + 1$ and divergent if $l \leq k + 1$.
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Exercise LXVII, problem 7, p. 311
If $m_{n}$ is a positive integer, and $m_{n+1} > m_{n}$, then the series $\sum r^{m_{n}}$ is convergent if $r < 1$ and divergent if $r \geq 1$. For example the series $1 + r + r^{4} + r^{9} + \dots$ is convergent if $r < 1$ and divergent if $r \geq 1$.
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Exercise LXVII, problem 8, p. 311
Sum the series $1 + 2r + 2r^{4} + \dots$ to $24$ places of decimals when $r = .1$ and to $2$ places when $r = .9$.
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Exercise LXVII, problem 9, p. 311
If $0 < a < b < 1$, then the series $a + b + a^{2} + b^{2} + a^{3} + \dots$ is convergent. Show that Cauchy’s test may be applied to this series, but that d’Alembert’s test fails.
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Exercise LXVIII

Exercise LXVIII, problem 1, p. 315
Verify that if $r < 1$ then 1 + r^2 + r + r^4 + r^6 + r^3 + … = 1 + r + r^3 + r^2 + r^5 + r^7 + … = 1/(1 - r).
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Exercise LXVIII, problem 2, p. 315
If either of the series $u_{0} + u_{1} + \dots$, $v_{0} + v_{1} + \dots$ is divergent, then so is the series $u_{0}v_{0} + (u_{1}v_{0} + u_{0}v_{1}) + (u_{2}v_{0} + u_{1}v_{1} + u_{0}v_{2}) + \dots$, except in the trivial case in which every term of one series is zero.
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Exercise LXVIII, problem 3, p. 315
If the series $u_{0} + u_{1} + \dots$, $v_{0} + v_{1} + \dots$, $w_{0} + w_{1} + \dots$ converge to sums $r$, $s$, $t$, then the series $\sum \lambda_{k}$, where $\lambda_{k} = \sum u_{m}v_{n}w_{p}$, the summation being extended to all sets of values of $m$, $n$, $p$ such that $m + n + p = k$, converges to the sum $rst$.
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Exercise LXVIII, problem 4, p. 315
If $\sum u_{n}$ and $\sum v_{n}$ converge to sums $s$ and $t$, then the series $\sum w_{n}$, where $w_{n} = \sum u_{l} v_{m}$, the summation extending to all pairs $l$, $m$ for which $lm = n$, converges to the sum $st$.
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Exercise LXIX

Exercise LXIX, problem 1, p. 317
Use Abel’s theorem to show that $\sum (1/n)$ and $\sum \{1/(an + b)\}$ are divergent. [Here $nu_{n} \to 1$ or $nu_{n} \to 1/a$.]
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Exercise LXIX, problem 2, p. 317
Show that Abel’s theorem is not true if we omit the condition that $u_{n}$ decreases as $n$ increases. [The series 1 + 12^2 + 13^2 + 14 + 15^2 + 16^2 + 17^2 + 18^2 + 19 + 110^2 + …, in which $u_{n} = 1/n$ or $1/n^{2}$, according as $n$ is or is not a perfect square, is convergent, since it may be rearranged in the form 12^2 + 13^2 + 15^2 + 16^2 + 17^2 + 18^2 + 110^2 + …+ (1 + 14 + 19 + …), and each of these series is convergent. But, since $nu_{n} = 1$ whenever $\DPtypo{u}{n}$ is a perfect square, it is clearly not true that $nu_{n} \to 0$.]
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Exercise LXIX, problem 3, p. 317
*The converse of Abel’s theorem is not true*, *i.e.* it is not true that, if $u_{n}$ decreases with $n$ and $\lim nu_{n} = 0$, then $\sum u_{n}$ is convergent. [Take the series $\sum(1/n)$ and multiply the first term by $1$, the second by $\frac{1}{2}$, the next two by $\frac{1}{3}$, the next four by $\frac{1}{4}$, the next eight by $\frac{1}{5}$, and so on. On grouping in brackets the terms of the new series thus formed we obtain 1 + 12 · 12 + 13 (13 + 14) + 14 (15 + 16 + 17 + 18) + …; and this series is divergent, since its terms are greater than those of 1 + 12 · 12 + 13 · 12 + 14 · 12 + …, which is divergent. But it is easy to see that the terms of the series 1 + 12 · 12 + 13 · 13 + 13 · 14 + 14 · 15 + 14 · 16 + … satisfy the condition that $nu_{n} \to 0$. In fact $nu_{n} = 1/\nu$ if $2^{\nu-2} < n \leq 2^{\nu-1}$, and $\nu \to \infty$ as $n \to \infty$.]
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Exercise LXX

Exercise LXX, problem 1, p. 319
Prove that _1^ 1n^2 + 1 < 12 + 14
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Exercise LXX, problem 2, p. 319
Prove that -12 < _1^ aa^2 + n^2 < 12 .
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Exercise LXX, problem 3, p. 319
Prove that if $m > 0$ then 1m^2 + 1(m + 1)^2 + 1(m + 2)^2 + … < m + 1m
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Exercise LXXI

Exercise LXXI, problem 1, p. 320
Prove by an argument similar to that used above, and without integration, that $\ds\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}}$, where $s < 1$, tends to infinity with $\xi$.
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Exercise LXXI, problem 2, p. 320
The series $\sum n^{-2}$, $\sum n^{-3/2}$, $\sum n^{-11/10}$ are convergent, and their sums are not greater than $2$, $3$, $11$ respectively. The series $\sum n^{-1/2}$, $\sum n^{-10/11}$ are divergent.
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Exercise LXXI, problem 3, p. 320
The series $\sum n^{s}/(n^{t} + a)$, where $a > 0$, is convergent or divergent according as $t > 1 + s$ or $t \leq 1 + s$. [Compare with $\sum n^{s-t}$.]
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Exercise LXXI, problem 4, p. 320
Discuss the convergence or divergence of the series (a_1n^s_1 + a_2n^s_2 + …+ a_kn^s_k)/ (b_1n^t_1 + b_2n^t_2 + …+ b_ln^t_l), where all the letters denote positive numbers and the $s$’s and $t$’s are rational and arranged in descending order of magnitude.
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Exercise LXXI, problem 5, p. 320
Prove that gather* 2n - 2 < 11 + 12 + …+ 1n < 2n - 1, 12 < 121 + 132 + 143 + … < 12(+ 1). gather* % [0]% (*Math. Trip.* 1911.)% [1]%
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Exercise LXXI, problem 6, p. 320
If $\phi(n) \to l > 1$ then the series $\sum n^{-\phi(n)}$ is convergent. If $\phi(n) \to l < 1$ then it is divergent.
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Exercise LXXII

Exercise LXXII, problem 1, p. 321
Show that if $a$ is any positive integer greater than $1$ then $\sum \phi(n)$ is convergent or divergent according as $\sum a^{n}\phi(a^{n})$ is convergent or divergent. [Use the same arguments as above, taking groups of $a$, $a^{2}$, $a^{3}$, … terms.]
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Exercise LXXII, problem 2, p. 321
If $\sum 2^{n}\phi(2^{n})$ converges then it is obvious that $\lim 2^{n}\phi(2^{n}) = 0$. Hence deduce Abel’s Theorem of [§]173.
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Exercise LXXIII

Exercise LXXIII, problem 1, p. 324
The integral _a^ x^r + x^r-1 + …+ Ax^s + Bx^s-1 + …+ L dx, where $\alpha$ and $A$ are positive and $a$ is greater than the greatest root of the denominator, is convergent if $s > r + 1$ and otherwise divergent.
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Exercise LXXIII, problem 10a, p. 324
**of Abel’s Theorem of [§]173.** *If $\phi(x)$ is positive and steadily decreases, and $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent, then $x\phi(x) \to 0$.* Prove this (*a*) by means of Abel’s Theorem and the Integral Test and (*b*) directly, by arguments analogous to those of [§]173.
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Exercise LXXIII, problem 10b, p. 324
**of Abel’s Theorem of [§]173.** *If $\phi(x)$ is positive and steadily decreases, and $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent, then $x\phi(x) \to 0$.* Prove this (*a*) by means of Abel’s Theorem and the Integral Test and (*b*) directly, by arguments analogous to those of [§]173.
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Exercise LXXIII, problem 11, p. 324
If $a = x_{0} < x_{1} < x_{2} < \dots$ and $x_{n} \to \infty$, and $\ds u_{n}= \int_{x_{n}}^{x_{n+1}} \phi(x)\, dx$, then the convergence of $\ds\int_{a}^{\infty} \phi(x)\, dx$ involves that of $\sum u_{n}$. If $\phi(x)$ is always positive the converse statement is also true. [That the converse is not true in general is shown by the example in which $\phi(x) = \cos x$, $x_{n} = n\pi$.]
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Exercise LXXIII, problem 2, p. 324
Which of the integrals %[** TN: All are displayed on one line in the original] $\ds\int_{a}^{\infty} \frac{dx}{\sqrt{x}}$, $\ds\int_{a}^{\infty} \frac{dx}{x^{4/3}}$, _a^ dxc^2 + x^2,0pt minus 3pt_a^ x dxc^2 + x^2,0pt minus 3pt_a^ x^2 dxc^2 + x^2,0pt minus 3pt_a^ x^2 dx+ 2x^2 + x^4 are convergent? In the first two integrals it is supposed that $a > 0$, and in the last that $a$ is greater than the greatest root (if any) of the denominator.
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Exercise LXXIII, problem 3, p. 324
The integrals _a^ x dx,0pt minus 3pt_a^ x dx,0pt minus 3pt_a^ (x + ) dx oscillate finitely as $\xi \to \infty$.
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Exercise LXXIII, problem 4, p. 324
The integrals _a^ xx dx,0pt minus 3pt_a^ x^2x dx0pt minus 3pt_a^ x^n (x + ) dx, where $n$ is any positive integer, oscillate infinitely as $\xi \to \infty$.
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Exercise LXXIII, problem 5, p. 324
**to $-\infty$.** If $\ds\int_{\xi}^{a} \phi(x)\, dx$ tends to a limit $l$ as $\xi \to -\infty$, then we say that $\ds\int_{-\infty}^{a} \phi(x)\, dx$ is convergent and equal to $l$. Such integrals possess properties in every respect analogous to those of the integrals discussed in the preceding sections: the reader will find no difficulty in formulating them.
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Exercise LXXIII, problem 6, p. 324
**from $-\infty$ to $+\infty$.** If the integrals _-^a (x) dx,0pt minus 3pt_a^ (x) dx are both convergent, and have the values $k$, $l$ respectively, then we say that _-^ (x) dx is convergent and has the value $k + l$.
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Exercise LXXIII, problem 7, p. 324
Prove that _-^0 dx1 + x^2 = _0^ dx1 + x^2 = 12 _-^ dx1 + x^2 = 12.
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Exercise LXXIII, problem 8, p. 324
Prove generally that _-^ (x^2) dx = 2_0^ (x^2) dx, provided that the integral $\ds\int_{0}^{\infty} \phi(x^{2})\, dx$ is convergent.
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Exercise LXXIII, problem 9, p. 324
Prove that if $\ds\int_{0}^{\infty} x\phi(x^{2})\, dx$ is convergent then $\ds\int_{-\infty}^{\infty} x\phi(x^{2})\, dx = 0$.
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Exercise LXXIV

Exercise LXXIV, problem 1, p. 327
Show, by means of the substitution $x = t^{\alpha}$, that if $s > 1$ and $\alpha >0$ then _1^ x^-s dx = _1^ t^(1-s) - 1 dt; and verify the result by calculating the value of each integral directly.
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Exercise LXXIV, problem 2, p. 327
If $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent then it is equal to one or other of _(a-)/^ (t + ) dt,0pt minus 3pt-_-^(a-)/ (t + ) dt, according as $\alpha$ is positive or negative.
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Exercise LXXIV, problem 3, p. 327
If $\phi(x)$ is a positive and steadily decreasing function of $x$, and $\alpha$ and $\beta$ are any positive numbers, then the convergence of the series $\sum \phi(n)$ implies and is implied by that of the series $\sum \phi(\alpha n + \beta)$.
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Exercise LXXIV, problem 4, p. 327
Show that %[** TN: In-line in the original] _1^ dx(1 + x)x = 12 .
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Exercise LXXIV, problem 5, p. 327
Show that _0^ x(1 + x)^2 dx = 12.
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Exercise LXXIV, problem 6, p. 327
If $\phi(x) \to h$ as $x \to \infty$, and $\phi(x) \to k$ as $x \to -\infty$, then _-^ (x - a) - (x - b) dx = -(a - b)(h - k).
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Exercise LXXV

Exercise LXXV, problem 1, p. 328
Show that %[** TN: In-line in the original] _0^ x(1 + x)^3 dx = 12 _0^ dx(1 + x)^2 = 12.
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Exercise LXXV, problem 2, p. 328
$\ds\int_{0}^{\infty} \frac{x^{2}}{(1 + x)^{4}}\, dx = \tfrac{2}{3} \int_{0}^{\infty} \frac{x}{(1 + x)^{3}}\, dx = \tfrac{1}{3}$.
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Exercise LXXV, problem 3, p. 328
If $m$ and $n$ are positive integers, and %[** TN: Two equations not displayed in the original] I_m, n = _0^ x^m dx(1 + x)^m+n, then I_m, n = m/(m + n - 1) I_m-1, n. Hence prove that $I_{m, n} = m!\, (n - 2)!/(m + n - 1)!$.
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Exercise LXXV, problem 4, p. 328
Show similarly that if %[** TN: Not displayed in the original] I_m, n = _0^ x^2m+1 dx(1 + x^2)^m+n then I_m, n = m/(m + n - 1) I_m-1, n,0pt minus 3pt2I_m, n = m! (n - 2)!/(m + n - 1)!. Verify the result by applying the substitution $x = t^{2}$ to the result of Ex. 3.
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Exercise LXXVI

Exercise LXXVI, problem 1, p. 331
If $\phi(x)$ is continuous except for $x = a$, while $\phi(x) \to \infty$ as $x \to a$, then the necessary and sufficient condition that $\ds\int_{a}^{A} \phi(x)\, dx$ should be convergent is that we can find a constant $K$ such that _a+^A (x) dx < K for all values of $\epsilon$, however small (cf. [§]178). It is clear that we can choose a number $A'$ between $a$ and $A$, such that $\phi(x)$ is positive throughout $\DPmod{(a, A')}{[a, A']}$. If $\phi(x)$ is positive throughout the whole interval $\DPmod{(a, A)}{[a, A]}$ then we can of course identify $A'$ and $A$. Now _a-^A (x) dx = _a-^A’ (x) dx + _A’^A (x) dx. The first integral on the right-hand side of the above equation increases as $\epsilon$ decreases, and therefore tends to a limit or to $\infty$; and the truth of the result stated becomes evident. If the condition is not satisfied then $\ds\int_{a-\epsilon}^{A} \phi(x)\, dx \to \infty$. We shall then say that the integral $\ds\int_{a}^{A} \phi(x)\, dx$ **** to $\infty$. It is clear that, if $\phi(x) \to \infty$ as $x \to a + 0$, then convergence and divergence to $\infty$ are the only alternatives for the integral. We may discuss similarly the case in which $\phi(x) \to -\infty$.
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Exercise LXXVI, problem 10, p. 331
Show that _0^h xx^p dx, where $0 < p < 2$, attains its greatest value when $h = \pi$. % [0]% (*Math. Trip.* 1911.)% [1]%
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Exercise LXXVI, problem 11, p. 331
The integral _0^12 (x)^l(x)^m dx is convergent if and only if $l > -1$, $m > -1$.
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Exercise LXXVI, problem 12, p. 331
Such an integral as _0^ x^s-1 dx1 + x, where $s < 1$, does not fall directly under any of our previous definitions. For the range of integration is infinite [pg]333 and the subject of integration tends to $\infty$ as $x \to +0$. It is natural to define this integral as being equal to the sum _0^1 x^s-1 dx1 + x + _1^ x^s-1 dx1 + x, provided that these two integrals are both convergent. 0.375em plus 0.75em minus 0.25emThe first integral is a convergent infinite integral of the second kind if $0 < s < 1$. The second is a convergent infinite integral of the first kind if $s < 1$. It should be noted that when $s > 1$ the first integral is an ordinary finite integral; but then the second is divergent. Thus the integral from $0$ to $\infty$ is convergent if and only if $0 < s < 1$.
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Exercise LXXVI, problem 13, p. 331
Prove that _0^ x^s-11 + x^t dx is convergent if and only if $0 < s < t$.
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Exercise LXXVI, problem 14, p. 331
The integral _0^ x^s-1 - x^t-11 - x dx is convergent if and only if $0 < s < 1$, $0 < t < 1$. [It should be noticed that the subject of integration is undefined when $x = 1$; but $(x^{s-1} - x^{t-1})/(1 - x) \to t - s$ as $x \to 1$ from either side; so that the subject of integration becomes a continuous function of $x$ if we assign to it the value $t - s$ when $x = 1$. It often happens that the subject of integration has a discontinuity which is due simply to a failure in its definition at a particular point in the range of integration, and can be removed by attaching a particular value to it at that point. In this case it is usual to suppose the definition of the subject of integration completed in this way. Thus the integrals _0^12 mxx dx,0pt minus 3pt_0^12 mxx dx are ordinary finite integrals, if the subjects of integration are regarded as having the value $m$ when $x = 0$.]
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Exercise LXXVI, problem 15, p. 331
**and integration by parts.** The formulae for transformation by substitution and integration by parts may of course be extended to infinite integrals of the second as well as of the first kind. The reader should formulate the general theorems for himself, on the lines of [§]179.
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Exercise LXXVI, problem 16, p. 331
Prove by integration by parts that if $s > 0$, $t > 1$, then _0^1 x^s-1(1 - x)^t-1 dx = t - 1s _0^1 x^s (1 - x)^t-2 dx.
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Exercise LXXVI, problem 17, p. 331
If $s > 0$ then _0^1 x^s-1 dx1 + x = _1^ t^-s dt1 + t. %[** TN: Added paragraph break] [Put $x = 1/t$.]
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Exercise LXXVI, problem 18, p. 331
If $0 < s < 1$ then _0^1 x^s-1 + x^-s1 + x dx = _0^ t^-s dt1 + t = _0^ t^s-1 dt1 + t.
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Exercise LXXVI, problem 19, p. 331
If $a + b > 0$ then _b^ dx(x + a)x - b = a + b. % [0]% (*Math. Trip.* 1909.)% [1]%
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Exercise LXXVI, problem 2, p. 331
Prove that _a^A (x - a)^-s dx = (A - a)^1-s1 - s if $s < 1$, while the integral is divergent if $s \geq 1$.
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Exercise LXXVI, problem 20, p. 331
Show, by means of the substitution $x = t/(1 - t)$, that if $l$ and $m$ are both positive then _0^ x^l-1(1 + x)^l+m dx = _0^1 t^l-1 (1 - t)^m-1 dt.
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Exercise LXXVI, problem 21, p. 331
Show, by means of the substitution $x = pt/(p + 1 - t)$, that if $l$, $m$, and $p$ are all positive then _0^1 x^l-1 (1 - x)^m-1 dx(x + p)^l + m = 1(1 + p)^l p^m _0^1 t^l-1 (1 - t)^m-1 dt.
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Exercise LXXVI, problem 22, p. 331
Prove that _a^b dx(x - a)(b - x) = 0pt minus 3ptand0pt minus 3pt_a^b x dx(x - a)(b - x) = 12 (a + b), (i) by means of the substitution $x = a + (b - a)t^{2}$, (ii) by means of the substitution $(b - x)/(x - a) = t$, and (iii) by means of the substitution $x = a\cos^{2} t + b\sin^{2} t$.
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Exercise LXXVI, problem 23, p. 331
If $s > -1$ then _0^12 ()^s d = _0^1 x^s dx1 - x^2 = 12 _0^1 x^12(s-1) dx1 - x = 12 _0^1 (1 - x)^12(s-1) dxx.
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Exercise LXXVI, problem 24, p. 331
Establish the formulae align* &_0^1 f(x) dx1 - x^2 = _0^12 f() d, % &_a^b f(x) dx(x - a)(b - x) = 2_0^12 f(a^2+ b^2) d, % &_-a^a fa - xa + x dx = 4a_0^12 f() d align*
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Exercise LXXVI, problem 25, p. 331
Prove that _0^1 dx(1 + x)(2 + x) x(1 - x) = (12 - 16) %[** Added paragraph break] [Put $x = \sin^{2}\theta$ and use % [examples:lxiii]Ex. lxiii%. 8.] % [0]% (*Math. Trip.* 1912.)% [1]% %[** TN: Dot added after "Math"]
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Exercise LXXVI, problem 3, p. 331
If $\phi(x) \to \infty$ as $x \to a + 0$ and $\phi(x) < K(x - a)^{-s}$, where $s < 1$, then $\ds\int_{a}^{A} \phi(x)\, dx$ is convergent; and if $\phi(x) > K(x - a)^{-s}$, where $s \geq 1$, then the integral is divergent. [This is merely a particular case of a general comparison theorem analogous to that stated in [§]178.]
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Exercise LXXVI, problem 4a, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4b, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4c, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4d, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4e, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4f, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 4g, p. 331
Are the integrals gather* _a^A dx(x - a)(A - x),0pt minus 3pt_a^A dx(A - x)[3]x - a,0pt minus 3pt_a^A dx(A - x)[3]A - x, _a^A dxx^2 - a^2,0pt minus 3pt_a^A dx[3]A^3 - x^3,0pt minus 3pt_a^A dxx^2 - a^2,0pt minus 3pt_a^A dxA^3 - x^3 gather* convergent or divergent?
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Exercise LXXVI, problem 5, p. 331
The integrals _-1^1dx[3]x,0pt minus 3pt_a-1^a+1 dx[3]x - a are convergent, and the value of each is zero.
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Exercise LXXVI, problem 6, p. 331
The integral _0^ dxx is convergent. [The subject of integration tends to $\infty$ as $x$ tends to either limit.]
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Exercise LXXVI, problem 7, p. 331
The integral _0^ dx(x)^s is convergent if and only if $s < 1$.
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Exercise LXXVI, problem 8, p. 331
The integral _0^12 x^s(x)^t dx is convergent if $t < s + 1$.
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Exercise LXXVI, problem 9, p. 331
Show that _0^h xx^p dx, where $h > 0$, is convergent if $p < 2$. Show also that, if $0 < p < 2$, the integrals _0^ xx^p dx,0pt minus 3pt_^2 xx^p dx,0pt minus 3pt_2^3 xx^p dx, … alternate in sign and steadily decrease in absolute value. [Transform the integral whose limits are $k\pi$ and $(k + 1)\pi$ by the substitution $x = k\pi + y$.]
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Exercise Misc-VIII

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