3.3 Series of variable terms. Uniformity of convergence.

Consider the series \[x^2 + \frac{x^2}{1+x^2}+ \frac{x^2}{(1+x^2)^2}+ \dots +\frac{x^2}{(1+x^2)^n}+ \dots .\] This series converges absolutely (§2.33) for all real values of \(x\).

If \(S_n (x)\) be the sum of \(n\) terms, then \[S_n(x)=1+x^2-\frac{1}{(1+x^2)^{n-1}} ;\] and so \[\lim_{n \rightarrow \infty} S_n(x)=1+x^2 ; \qquad (x \neq 0)\] but \(S_n (0) = 0\), and therefore \(\lim\limits_{n \rightarrow \infty} S_n (0) = 0\).

Consequently, although the series is an absolutely convergent series of continuous functions of \(x\), the sum is a discontinuous function of \(x\). We naturally enquire the reason of this rather remarkable phenomenon, which was investigated in 1841–1848 by Stokes[1], Seidel[2] and Weierstrass[3], who shewed that it cannot occur except in connexion with another phenomenon, that of non-uniform convergence, which will now be explained.

[1]Camb. Phil. Trans, viii. (1847), pp. 533–583. [Collected Papers, i. pp. 236–313.]  ↩
[2]Münchener Abhandlungen, v. (1848), p. 381.  ↩
[3]Ges. Math. Werke, i. pp. 67, 75.  ↩

Let the functions \(u_1(z),\, u_2(z), \dots \) be defined at all points of a closed region of the Argand diagram. Let \[S_n(x)=u_1(z)+u_2(z)+ \dots +u_n(z) .\]

The condition that the series \(\sum\limits_{n=1}^{\infty} u_n (z)\) should converge for any particular value of \(z\) is that, given \(\epsilon\), a number \(n\) should exist such that \[\left|\,S_{n+p}(z) - S_n(z) \,\right| < \epsilon\] for all positive values of \(p\), the value of \(n\) of course depending on \(\epsilon\).

Let \(n\) have the smallest integer value for which the condition is satisfied. This integer will in general depend on the particular value of \(z\) which has been selected for consideration. We denote this dependence by writing \(n(z)\) in place of \(n\). Now it may happen that we can find a number \(N\), independent of \(z\), Such that \[n(x) < N\] for all values of \(z\) in the region under consideration.

If this number \(N\) exists, the series is said to converge uniformly throughout the region.

If no such number \(N\) exists, the convergence is said to be non-uniform.[4]

[4]The reader who is unacquainted with the concept of uniformity of convergence will find it made much clearer by consulting Bromwich, Infinite Series, Ch. vii, where an illuminating account of Osgood’s graphical investigation is given. ↩

Uniformity of convergence is thus a property depending on a whole set of values of \(z\), whereas previously we have considered the convergence of a series for various particular values of \(z\), the convergence for each value being considered without reference to the other values.

We define the phrase ‘uniformity of convergence near a point \(z\)’ to mean that there is a definite positive number \(\delta\) such that the series converges uniformly in the domain common to the circle \(\left|\, z - z_1 \right | \leq \delta\) and the region in which the series converges.

3.31 On the condition for uniformity of convergence.[5]

[5]This section shews that it is indifferent whether uniformity of convergence is defined by means of the partial remainder \(R_{n,\:\! p}(z)\) or by \(R_n (z)\). Writers differ in the definition taken as fundamental. ↩

If \(R_{n, \:\! p}(z) = u_{n+1} (z) + u_{n+2} (z)+ \dots + u_{n+p} (z)\), we have seen that the necessary and sufficient condition that \(\sum_{n=1}^{\infty}u_n (z)\) should converge uniformly in a region is that, given any positive number \(\epsilon\), it should be possible to choose \(N\) independent of \(z\) (but depending on \(\epsilon\)) such that \[\left|\, R_{N, \:\! p}(z)\,\right| < \epsilon\] for all positive integral values of \(p\).

If the condition is satisfied, by §2.22, \(S_n (z)\) tends to a limit, \(S(z)\), say for each value of \(z\) under consideration; and then, since \(\epsilon\) is independent of \(p\), \[\left|\, \lim_{p \rightarrow \infty} R_{N, \:\! p} (z) \,\right| \leq \epsilon ,\] and therefore, when \(n > N\), \[S(z)-S_n(z)= \lim_{p \rightarrow \infty} R_{N, \:\! p} (z)-R_{N, \:\! n-N}(z) ,\] and so \[\left|\, S(z)-S_n(z)\,\right| < 2\epsilon .\]

Thus (writing \(\frac{1}{2} \epsilon\) for \(\epsilon\) ) a necessary condition for uniformity of convergence is that \(\left|\,S(z) - S_n (z) \,\right| < \epsilon,\) whenever \(n>N\) and \(N\) is independent of \(z\); the condition is also sufficient; for if it is satisfied it follows as in §2.22(I) that \(\left|\, R_{N,\:\! p} (z)\, \right| < 2\epsilon\), which, by definition, is the condition for uniformity.

Example 1. Shew that, if \(x\) be real, the sum of the series \[\frac{x}{1(x+1)}+\frac{x}{(x+1)(2x+1)}+ \cdots +\frac{x}{[(n-1)x+1](nx+1)}+ \cdots\] is discontinuous at \(x=0\) and the series is non-uniformly convergent near \(x=0.\)

The sum of the first \(n\) terms is easily seen to be \(\displaystyle{1-\frac{1}{nx+1}}\); so when \(x=0\) the sum is \(0\); when \(x \neq 0\), the sum is \(1\).

The value of \(R_n(x) = S(x)-S_n (x)\) is \(\displaystyle{\frac{1}{nx+1}}\) if \(x \neq 0\); so when \(x\) is small, say \(x=\)one-hundred-millionth, the remainder after a million terms is \(\frac{1}{\frac{1}{100}+1}\) or \(1-\frac{1}{101}\), so the first million terms of the series do not contribute one percent of the sum. And in general, to make \(\displaystyle{\frac{1}{nx+1}}<\epsilon\), it is necessary to take \[n>\frac{1}{x}\left( \frac{1}{\epsilon}-1\right).\]

Corresponding to a given \(\epsilon\), no number \(N\) exists, independent of \(x\), such that \(n < N\) for all values of \(x\) in any interval including \(x=0\); for by taking \(x\) sufficiently small we can make \(n\) greater than any number \(N\) which is independent of \(x\). There is therefore non-uniform convergence near \(x = 0.\)

Example 2. Discuss the series \[\sum_{n=1}^{\infty} \frac{x\{n(n+1)x^2-1\}}{( 1+n^2 x^2 )(1+(n+1)^2 x^2)},\] in which \(x\) is real.

The \(n\)th term can be written \(\: \displaystyle \frac{nx}{1+n^2x^2} - \frac{(n+1)x}{1+(n+1)^2x^2}, \: \) so \(\: \displaystyle S(x) =\frac{x}{1+x^2},\: \) and \[R_n(x)=\frac{(n+1)x}{1+(n+1)^2x^2}.\]

[Note. In this example the sum of the series is not discontinuous at \(x=0\).]

But (taking \(\epsilon <\frac{1}{2}\) and \(x \neq 0\) ), \(\left|\, R_n (x) \,\right| <\epsilon \) if’ \(\epsilon^{-1}(n+1)\,\left|\,x\,\right| < 1+(n+1)^2 x^2\); i.e. if \(n+1>\frac{1}{2}\{\epsilon^{-1}+\sqrt{\epsilon^{-2}-4}\,\}\,\left|\,x\,\right|^{-1}\) or \(n+1<\frac{1}{2}\{\epsilon^{-1}-\sqrt{\epsilon^{-2}-4}\,\}\,\left|\,x\,\right|^{-1}\).[6]

[6]Editor’s Note: Rewrite \(\epsilon^{-1}(n+1)\,\left|\,x\,\right| < 1+(n+1)^2 x^2\) as \((n+1)^2 x^2-\epsilon^{-1}(n+1)\,\left|\,x\,\right| + 1>0\), thinking of the left hand side as a quadratic in \((n+1)\,\left|\,x\,\right|\), and solve. ↩

Now it is not the case that the second inequality is satisfied for all values of \(n\) greater than a certain value and for all values of \(x\); and the first inequality gives a value of \(n(x)\) which tends to infinity as \(x \rightarrow 0 \); so that, corresponding to any interval containing the point \(x=0\), there is no number \(N\) independent of x. The series, therefore, is non-uniformly convergent near \(x=0\).

The reader will observe that \(n(x)\) is discontinuous at \(x = 0\); for \(n(x) \rightarrow \infty\) as \(x \rightarrow 0 \), but \(n(0)=0\).

3.32 Connexion of discontinuity with non-uniform convergence.

We shall now shew that if a series of continuous functions of \(z\) is uniformly convergent for all values of \(z\) in a given closed domain, the sum is a continuous function of \(z\) at all points of the domain.

For let the series be \(f(z) = u_1 (z) + u_2 (z) + \cdots + u_n (z)+ \cdots = S_n (z) + R_n (z)\), where \(R_n (z)\) is the remainder after \(n\) terms.

Since the series is uniformly convergent, given any positive number \(\epsilon\), we can find a corresponding integer \(n\) independent of \(z\), such that \(\left|\, R_n (z) \,\right| < \frac{1}{3} \epsilon \) for all values of \(z\) within the domain.

Now \(n\) and \(\epsilon\) being thus fixed, we can, on account of the continuity of \(S_n(z)\), find a positive number \(\eta\) such that \[ \left|\, S_n(z)-S_n (z')\,\right|< \frac{1}{3}\epsilon,\] whenever \(\left|\, z — z' \, \right| < \eta\).

We have then \[ \begin{align*} \left|\,f(z)-f(z')\, \right|& \leq \left|\,S_n(z)-S_n(z') \, \right|+ \left|\,R_n(z)-R_n(z') \, \right|\\ \\ & \leq \left|\,S_n(z)-S_n(z') \, \right|+ \left|\,R_n(z) \,\right|+ \left|\,R_n(z') \, \right|\\ \\ & < \epsilon , \end{align*} \] which is the condition for continuity at \(z\).

Example 1. Shew that near \(x = 0\) the series \[u_1 (x) + u_2 (x) + u_3 (x)+ \cdots,\] \[\text{where} \qquad u_1(x) = x,\quad u_n (x) = x^{\left.1\middle/(2n - 1)\right.} - x^{\left.1\middle/(2n - 3)\right.}, \] and real values of \(x\) are concerned, is discontinuous and non-uniformly convergent.

In this example it is convenient to take a slightly different form of the test; we shall shew that, given an arbitrarily small number \(\epsilon\), it is possible to choose values of \(x\), as small as we please, depending on \(n\) in such a way that \(\left|\, R_n (x) \,\right|\) is not less than \(\epsilon\) for any value of \(n\), no matter how large. The reader will easily see that the existence of such values of \(x\) is inconsistent with the condition for uniformity of convergence.

The value of \(S_n (x)\) is \(x^{\left.1\middle/(2n - 1)\right.}\); as \(n\) tends to infinity, \(S_n (x)\) tends to \(1\), \(0\), or \(-1\), according as \(x\) is positive, zero, or negative. The series is therefore absolutely convergent for all values of \(x\), and has a discontinuity at \(x=0\).

In this series \(R_n (x) = 1 - x^{\left.1\middle/(2n - 1)\right.}\), \((x > 0)\); however great \(n\) may be, by taking[7] \(x = \epsilon^{-(2n - 1)}\) we can cause this remainder to take the value \(1-\epsilon^{-1}\), which is not arbitrarily small. The series is therefore non-uniformly convergent near \(x=0\).

[7]This value of \(x\) satisfies the condition \(\left|\,x\,\right| < \delta \) whenever \(2n - 1 > \log \delta^{\:\! -1}\). ↩

Example 2. Shew that near \(z=0\) the series \[\sum_{n=1}^{\infty} \frac{-2z(1+z)^{n-1}}{\{1+(1+z)^{n-1}\}\{1+(1+z)^{n}\}}\] is non-uniformly convergent and its sum is discontinuous.

The \(n\)th term can be written \[\frac{1-(1+z)^{n}}{1+(1+z)^{n}}-\frac{1-(1+z)^{n-1}}{1+(1+z)^{n-1}},\] so the sum of the first \(n\) terms is \[\frac{1-(1+z)^{n}}{1+(1+z)^{n}}.\] Thus, considering real values of \(z\) greater than –1, it is seen that the sum to infinity is 1, 0, or –1, according as \(z\) is negative, zero, or positive. There is thus a discontinuity at \(z=0\). This discontinuity is explained by the fact that the series is non-uniformly convergent near \(z=0\); for the remainder after \(n\) terms in the series when \(z\) is positive is \[\frac{-2}{1+(1+z)^{n}},\] and, however great \(n\) may be, by taking \(z=\frac{1}{n}\), this can be made numerically greater than \(\frac{2}{1+e},\) which is not arbitrarily small. The series is therefore non-uniformly convergent near \(z=0\).

3.33 The distinction between absolute and uniform convergence.

The uniform convergence of a series in a domain does not necessitate its absolute convergence at any points of the domain, nor conversely. Thus the series \(\sum \frac{z^2}{1+z^2}\), converges absolutely, but (near \(z = 0\)) not uniformly; while in the case of the series \[\sum_{n=1}^\infty \frac{(-1)^{n-1}}{z^2+n},\] the series of moduli is \[\sum_{n=1}^\infty \frac{1}{\left|\, n+z^2 \,\right|},\] which is divergent, so the series is only conditionally convergent; but for all real values of \(z\), the terms of the series are alternately positive and negative and numerically decreasing, so the sum of the series lies between the sum of its first \(n\) terms and of its first \((n + 1)\) terms, and so the remainder after \(n\) terms is numerically less than the \(n\)th term. Thus we only need take a finite number (independent of \(z\),) of terms in order to ensure that for all real values of \(z\), the remainder is less than any assigned number \(\epsilon\) and so the series is uniformly convergent.

Absolutely convergent series behave like series with a finite number of terms in that we can multiply them together and transpose their terms.

Uniformly convergent series behave like series with a finite number of terms in that they are continuous if each term in the series is continuous and (as we shall see) the series can then be integrated term by term.

3.34 A condition, due to Weierstrass,[8] for uniform convergence.

[8]Abhandlungen aus der Funktionerilehre, p. 70. The test given by this condition is usually described (e.g. by Osgood, Annals of Mathematics, iii. (1901), p. 130) as the M-test. ↩

A sufficient, though not necessary, condition for the uniform convergence of a series may be enunciated as follows:

If, for all values of \(z\) within a domain, the moduli of the terms of a series \[S=u_1(z)+u_2(z)+u_3(z)+ \cdots\] are respectively less than the corresponding terms in a convergent series of positive terms \[T=M_1 + M_2 + M_3 + \cdots ,\] where \(M_n\) is independent of \(z\), then the series \(S\) is uniformly convergent in this region. This follows from the fact that, the series \(T\) being convergent, it is always possible to choose \(n\) so that the remainder after the first \(n\) terms of \(T\), and therefore the modulus of the remainder after the first \(n\) terms of \(S\), is less than an assigned positive number \(\epsilon\); and since the value of \(n\) thus found is independent of \(z\), it follows (§3.31) that the series \(S\) is uniformly convergent; by §2.34, the series \(S\) also converges absolutely.

Example. The series \[\cos z +\frac{1}{2^2}\cos^2 z +\frac{1}{3^2}\cos^3 z + \cdots\] convergent for all real values of \(z\), because the moduli of its terms are not greater than the corresponding terms of the convergent series \[1 +\frac{1}{2^2} +\frac{1}{3^2} + \cdots ,\] whose terms are positive constants.

3.341 Uniformity of convergence of infinite products.[9]
[9]The definition is, effectively, that given by Osgood, Funktionentheorie, p. 462. The condition here given for uniformity of convergence is also established in that work. ↩

A convergent product \(\prod\limits_{n=1}^\infty \{1+u_n(z)\}\) is said to converge uniformly in a domain of values of \(z\) if, given \(\epsilon,\) we can find \(m\) independent of \(z\) such that \[\left|\,\prod\limits_{n=1}^{m+p} \{1+u_n(z)\}-\prod\limits_{n=1}^m \{1+u_n(z)\} \,\right| < \epsilon\] for all positive integral values of \(p\).

The only condition for uniformity of convergence which will be used in this work is that the product converges uniformly if \(\left|\, u_n (z) \,\right| < M_n\) where \(M_n\) is independent of \(z\) and \(\sum\limits_{n=1}^\infty M_n\) converges.

To prove the validity of the condition we observe that \(\prod\limits_{n=1}^\infty (1+M_n)\) converges (§2.7), and so we can choose \(m\) such that \[\prod\limits_{n=1}^{m+p} \{1+M_n\}-\prod\limits_{n=1}^m \{1+M_n\} < \epsilon ;\] and then we have \[ \left|\,\prod\limits_{n=1}^{m+p} \{1+u_n(z)\}-\prod\limits_{n=1}^m \{1+u_n(z)\} \,\right| \] \[ \begin{align*} \qquad \qquad \quad & =\left|\, \prod\limits_{n=1}^m \{1+u_n(z)\} \left[ \,\prod\limits_{n=m+1}^{m+p} \{1+u_n(z)\}-1 \,\right] \,\right|\\ & \leq \, \prod\limits_{n=1}^m \{1+M_n\} \left[ \,\prod\limits_{n=m+1}^{m+p} \{1+M_n\}-1 \,\right] \\ \\ \\ & < \epsilon , \end{align*} \] and the choice of \(m\) is independent of \(z\).

3.35 Hardy’s Tests for Uniform Convergence.[10]

[10]Proc. London Math. Soc. (2) iv. (1907), pp. 247–265. These results, which are generalisations of Abel’s theorem (§3.71, below), though well known, do not appear to have been published before 1907. From their resemblance to the tests of Dirichlet and Abel for convergence, Bromwich proposes to call them Dirichlet’s and Abel’s tests respectively. ↩

The reader will see, from §2.31, that if, in a given domain, \(\left|\,\sum_{n=1}^p a_b(z)\,\right| < k\) where \( a_n (z)\) is real and \(k\) is finite and independent of \(p\) and \(z\), and if \(f_n (z) \geq f_{n + 1} (z)\) and \( f_n(z) \rightarrow 0\) uniformly as \( n \rightarrow \infty\), then \(\sum\limits_{n=1}^\infty a_n(z) \, f_n(z)\) converges uniformly.

Also that if \[k \geq u_n(z) \geq u_{n+1}(z) \geq 0 ,\] where \(k\) is independent of \(z\) and \( \sum\limits_{n=1}^\infty a_n (z)\) converges uniformly, then \( \sum\limits_{n=1}^\infty a_n (z)\,u_n (z)\) converges uniformly. [To prove the latter, observe that \(m\) can be found such that \[ a_{m+1}(z), \; a_{m+1}(z)+a_{m+2}(z), \cdots , \; a_{m+1}(z)+a_{m+2}(z)+ \cdots+ a_{m+p}(z) \] are numerically less than \( \left.\epsilon \middle/ k\right.\); and therefore (§2.301) \[\left|\,\sum_{n=m+1}^{m+p} a_n(z)\:\! u_n(z)\,\right| < \frac{\epsilon u_{m+1}(z)}{k} < \epsilon ,\] and the choice of \(\epsilon\) and \(m\) to is independent of \(z\).]

Example 1. Shew that, if \(\delta >0\), the series \[\sum_{n=1}^\infty \frac{\cos n\theta}{n}, \quad \sum_{n=1}^\infty \frac{\sin n\theta}{n}\] converge uniformly in the range \(\delta \leq \theta \leq 2\pi-\delta\).

Obtain the corresponding result for the series \[\sum_{n=1}^\infty \frac{ (-1)^n\cos n\theta}{n}, \quad \sum_{n=1}^\infty \frac{ (-1)^n\sin n\theta}{n}.\] by writing \(\theta +\pi\) for \(\pi\).

Example 2. If, when \( a \leq x \leq b\), \(\left|\,\omega_n (x)\, \right| < k_1\) and \(\sum\limits_{n=1}^\infty \left|\, \omega_{n+1}(x)-\omega_n(x) \, \right| < k_2\), where \(k_1, \, k_2\) are independent of \(n\) and \(x\), and if \(\sum\limits_{n=1}^\infty a_n\) is a convergent series independent of \(x\), then \(\sum\limits_{n=1}^\infty a_n \:\!\omega_n(x)\) converges uniformly when \( a \leq x \leq b\).
(Hardy.)

[For we can choose \(m\), independent of \(x\), such that \(\left|\, \sum_{n=m+1}^{m+p} a_n \,\right| < \epsilon\) and then, by §2.301 corollary, we have \(\left|\, \sum_{n=m+1}^{m+p} a_n \:\!\omega_n (x) \,\right| < (k_1+k_2)\epsilon.\) ]

3.4 Discussion of a particular double series.

Let \(\omega_1\) and \(\omega_2\) be any constants whose ratio is not purely real; and let \(\alpha\) be positive.

The series \[\sum\limits_{m,\,n} \frac{1}{(z+2m\omega_1+2n\omega_2)^\alpha},\] in which the summation extends over all positive and negative integral and zero values of \(m\) and \(n\), is of great importance in the theory of Elliptic Functions. At each of the points \(z = - 2m\omega_1-2n\omega_2\) the series does not exist. It can be shewn that the series converges absolutely for all other values of \( z\) if \(\alpha > 2\), and the convergence is uniform for those values of \(z\) such that \(\left|\,z+2m\omega_1+2n\omega_2 \,\right| \geq \delta\) for all integral values of \(m\) and \(n\), where \(\delta\) is an arbitrary positive number.

Let \(\sum^\prime_{m,\,n}\) denote a summation for all integral values of \(m\) and \(n\), the term for which \(m = n = 0\) being omitted.

Now, if \(m\) and \(n\) are not both zero, and if \(\left|\,z+2m\omega_1+2n\omega_2 \,\right| \geq \delta > 0\) for all integral values of \(m\) and \(n\), then we can find a positive number \(C\). depending on \(\delta\) but not on \( z\), such that \[\left| \, \frac{1}{(z+2m\omega_1+2n\omega_2)^\alpha} \,\right| < C\,\left| \, \frac{1}{(2m\omega_1+2n\omega_2)^\alpha} \,\right|.\]

Consequently, by §3.34, the given series is absolutely and uniformly[11] convergent in the domain considered if \[\sum_{m,\,n}' \frac{1}{\left|\, m\omega_1+n\omega_2 \right|^\alpha}\] converges.

[11]The reader will easily define uniformity of convergence of double series (see §3.5).  ↩

To discuss the convergence of the latter series, let \[\omega_1=\alpha_1+i\beta_1, \quad \omega_2=\alpha_2+i\beta_2,\] where \(\alpha_1,\,\alpha_2,\;\beta_1, \,\beta_2\) are real. Since \(\\\omega_1 \left/ \omega_2 \right.\) is not real, \(\alpha_1\beta_2-\alpha_2\beta_1 \neq 0\). Then the series is \[\sum_{m,\,n}' \frac{1}{\{(\alpha_1 m+\alpha_2 n)^2+(\beta_1 m+\beta_2 n)^2\}^{\left.\alpha\middle/2\right.}}.\] This converges if the series \[\sum_{m,\,n}' \frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}}\] converges; for the quotient of corresponding terms is \[\left\{\frac{(\alpha_1+\alpha_2 \mu)^2+(\beta_1+\beta_2 \mu)^2}{1+\mu^2} \right\}^{\left.\alpha\middle/2\right.}\] where \(\mu = n \left/m \right.\). This expression, qua[12] function of a continuous real variable \(\mu\), can be proved to have a positive minimum[13] (not zero) since \(\alpha_1\beta_2-\alpha_2\beta_1 \neq 0\) and so the quotient is always greater than a positive number \(K\) (independent \(\mu\)).[14]

[12]Editor’s Note: In this usage, the word qua means functioning as a…  ↩
[13]The reader will find no difficulty in verifying this statement; the minimum value in question is given by \[\begin{align*}K^{2/\alpha}&=\frac{1}{2}\left[ \,\alpha_1^2+\alpha_2^2+\beta_1^2+\beta_2^2 \right.\\&-\left\{(\alpha_1-\beta_2)^2+(\alpha_2+\beta_1)^2\right\}^{1/2} \\&\times \left.\left\{(\alpha_1+\beta_2)^2+(\alpha_2-\beta_1)^2\right\}^{1/2}\right].\end{align*}\] ↩
[14]Editor’s Note: For clarity’s sake, we just demonstrated that \[\begin{align*}&\frac{1}{(m^2+n^2)^{\frac{1}{2}\alpha}} \geq\\&\frac{K}{\{(\alpha_1 m+\alpha_2 n)^2+(\beta_1 m+\beta_2 n)^2\}^{\frac{1}{2}\alpha}}.\end{align*}\] ↩

We have therefore only to study the convergence of the series \(S\). Let \[ \begin{align*} S_{p,\,q} &=\sum_{m=-p}^p {\sum_{n=-q}^q}\!^\prime \frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}} \\ & \leq 4 \sum_{m=0}^p {\sum_{n=0\;}^q}\!^\prime\frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}}. \end{align*} \]

Separating \(S_{p,\,q}\) into the terms for which \(m = n\), \(m > n\), and \(m < n\), respectively, we have \[\frac{1}{4} S_{p,\,q} = \sum_{m=1}^p \frac{1}{(2m^2)^{\left.\alpha\middle/2\right.}}+\sum_{m=1}^p \sum_{n=0}^{m-1}\frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}}+\sum_{n=1}^p \sum_{m=0}^{n-1}\frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}}.\] But \[\sum_{n=0}^{m-1}\frac{1}{(m^2+n^2)^{\left.\alpha\middle/2\right.}} < \frac{m}{(m^2)^{\left.\alpha\middle/2\right.}}=\frac{1}{m^{\alpha-1}}.\] Therefore \[\frac{1}{4} S \leq \sum_{m=1}^\infty \frac{1}{2^{\left.\alpha\middle/2\right.}m^\alpha}+\sum_{m=1}^\infty \frac{1}{m^{\alpha-1}}+\sum_{n=1}^\infty \frac{1}{m^{\alpha-1}}.\] But these last series are known to be convergent if \(\alpha -1 > 1\). So the series \(S\) is convergent if \(\alpha > 2\). The original series is therefore absolutely and uniformly convergent, when \(\alpha > 2\), for the specified range of values of \(z\).

Example. Prove that the series \[\sum \frac{1}{(m_1^2+m_2^2+ \cdots +m_r^2)^\mu}\] in which the summation extends over all positive and negative integral values and zero values of \(m_1,\, m_2, \dots ,\,m_r,\) except the set of simultaneous zero values, is absolutely convergent if \(\mu > \frac{1}{2}r\).
(Eisenstein, Journal für Math., xxxv. )