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I am now going to share a solution on $$S=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n^{2}+k^2}$$ using the Theorem $\displaystyle \pi \csc (\pi z)=\lim _{N \rightarrow \infty} \sum_{n=-N}^{N} \frac{(-1)^{k}}{k+z},\tag*{(*)} \textrm{ where } z \not \in Z. $

We first split the series into 2 by partial fractions.

$\displaystyle \begin{aligned}S &=-\frac{1}{2 ki } \sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{n+ki}-\frac{(-1)^{n}}{n-ki}\right) \\&=-\frac{1}{2 ki}\left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+ki}-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n-ki}\right) \end{aligned}\tag*{} $

Re-indexing n to convert $S$ back to a single series yields

$\displaystyle \begin{aligned}S&=-\frac{1}{2 ki}\left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+ki}+\sum_{n=-1}^{-\infty} \frac{(-1)^{n}}{n+ki}+\frac{1}{ki} \right) \\&=-\frac{1}{2 ki}\left(\sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{n+ki}+\frac{1}{ki}\right)\end{aligned}\tag*{} $ Applying the Theorem (*) gives $\displaystyle \begin{aligned}S &=-\frac{1}{2 ki}\left(\pi \csc (k\pi i)+\frac{1}{ki}\right) \\&=-\frac{1}{2 ki}\left(-i \pi \operatorname{csch}(k\pi)+\frac{1}{ki}\right)\\&=\frac{1}{2k}\left(\pi \operatorname{csch} (k\pi)+\frac{1}{2k} \right)\\& \left(\textrm{ OR } \frac{1}{4k^2}+\frac{\pi}{k(e^{\pi}-e^{-\pi})}\right)\end{aligned}\tag*{} $

Your comments and alternative methods are highly appreciated.

Lai
  • 20,421

1 Answers1

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After the partial fraction, computing the sums in terms of the polygamma function, converting to generalized harmonic numbers give $$S_p=\sum_{n=0}^{p} \frac{(-1)^{n}}{n^{2}+k^2}=\frac{1+\pi k \coth (\pi k)-i k \left(H_{p-i k}-H_{p+i k}\right)}{2 k^2}$$ Using the asymptotics $$S_p=\frac{1+\pi k \coth (\pi k)}{2 k^2}-\frac{1}{p}+\frac{1}{2 p^2}+\frac{2 k^2-1}{6 p^3}+O\left(\frac{1}{p^4}\right)$$

Claude Leibovici
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