Using complex analysis, it is well known, that $$f(s)=\sum_{k=1}^\infty \frac{2s^2}{k^2+s^2} = \pi s \coth(\pi s) - 1 \, .$$ From this it is clear, that $$f\left(i(n+1/2)\right) = \sum_{k=1}^\infty \frac{-2(2n+1)^2}{(2k)^2-(2n+1)^2} = -1 \, . \tag{1}$$ Similarly $$g(s)=\sum_{k=0}^\infty \frac{2s}{(2k+1)^2-s^2} = \frac{\pi}{2} \, \tan\left(\frac{\pi s}{2}\right)$$ and so $$g(2n)=\sum_{k=0}^\infty \frac{4n}{(2k+1)^2-(2n)^2} = 0\,. \tag{2}$$ Proving these seems to require complex analysis. Is it possible to prove (1) and (2) using only elementary methods?
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Perform partial fraction decomposition, and note that, although the harmonic series diverges, its tails vanish when the tail length is fixed: $$\begin{align}\sum_{k\ge0}\frac{4n}{(2k+1)^2-(2n)^2}&=\sum_{k\ge0}\frac{4n}{(2k-2n+1)(2k+2n+1)}\\&=\sum_{k\ge0}\left[\frac{1}{2k-2n+1}-\frac{1}{2k+2n+1}\right]\\&=\lim_{N\to\infty}\left[\sum_{k=0}^N\frac{1}{2(k-n)+1}-\sum_{k=2n}^{N+2n}\frac{1}{2(k-n)+1}\right]\\&=\lim_{N\to\infty}\left[\sum_{k=0}^{2n-1}\frac{1}{2(k-n)+1}-\sum_{k=N+1}^{N+2n}\frac{1}{2(k-n)+1}\right]\\&=\sum_{k=0}^{2n-1}\frac{1}{2(k-n)+1}\\&=1+\sum_{k=0}^{n-1}\frac{1}{2(k-n)+1}+\sum_{k=n+1}^{2n-1}\frac{1}{2(k-n)+1}\\&=1+\sum_{k=-n}^{-1}\frac{1}{2k+1}+\sum_{k=1}^{n-1}\frac{1}{2k+1}\\&=1+\sum_{k=1}^n\frac{1}{1-2k}+\sum_{k=2}^n\frac{1}{2k-1}\\&=1+\frac{1}{1-2(1)}+0\\&=0\end{align}$$
And similarly (using the $0$ sum from the first part):
$$\begin{align}\sum_{k\ge1}\frac{-2(2n+1)^2}{(2k)^2-(2n+1)^2}&=\sum_{k\ge1}(2n+1)\left[\frac{1}{2k+2n+1}-\frac{1}{2k-2n-1}\right]\\&=\lim_{N\to\infty}\left[\sum_{k=2n+2}^{N+2n+1}\frac{2n+1}{2(k-n)-1}-\sum_{k=1}^N\frac{2n+1}{2(k-n)-1}\right]\\&=\lim_{N\to\infty}\left[\sum_{k=N+1}^{N+2n+1}\frac{2n+1}{2(k-n)-1}-\sum_{k=1}^{2n+1}\frac{2n+1}{2(k-n)-1}\right]\\&=-(2n+1)\cdot\sum_{k=0}^{2n}\frac{1}{2(k-n)+1}\\&=-(2n+1)\cdot\left(0+\frac{1}{2(2n-n)+1}\right)\\&=-1\end{align}$$
0
$$\frac{1}{(2k+1)^2-(2n)^2}=\frac{1}{4 n (2 k-2n+1)}-\frac{1}{4 n (2k+2 n+1)}$$ $$S_1=\sum_{k=0}^p \frac{1}{2 k-2n+1}=\frac{1}{2} \left(H_{p-n+\frac{1}{2}}-H_{-n-\frac{1}{2}}\right)$$ $$S_2=\sum_{k=0}^p \frac{1}{2 k+2n+1}=\frac{1}{2} \left(H_{p+n+\frac{1}{2}}-H_{n-\frac{1}{2}}\right)$$
$S_1-S_2$ simplifies $$S_1-S_2=\frac{1}{2} \left(H_{p-n+\frac{1}{2}}-H_{p+n+\frac{1}{2}}+\pi \tan (\pi n)\right)$$ So, if $n$ is an integer $$S_1-S_2=\frac{1}{2} \left(H_{p-n+\frac{1}{2}}-H_{p+n+\frac{1}{2}}\right)$$
Using the asymptotics of harmonic numbers $$S_1-S_2=-\frac{n}{p}+\frac{n}{p^2}-\frac{n \left(4 n^2+11\right)}{12 p^3}+O\left(\frac{1}{p^4}\right)$$ $$\frac{S_1-S_2}{4n}=-\frac{1}{4 p}+\frac{1}{4 p^2}-\frac{4 n^2+11}{48 p^3}+O\left(\frac{1}{p^4}\right)=-\frac{1}{4 p}+O\left(\frac{1}{p^2}\right)$$
Claude Leibovici
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