TITLE: How to find $\sum_{k \in \mathbb{Z}}\frac1{(k+a)(k+b)}$
QUESTION [5 upvotes]: Let $a,b$ be two unequal integers. I have to find the sum below.
$$
\sum_{k \in \mathbb{Z}}\frac1{(k+a)(k+b)}
$$
I should use complex analysis, but I have no clue where to start. I only now that I can assume without loss of generality that $a = 0$ because $x \mapsto x -a$ is an invertible map on the set of integers. I would like to get I hint from you, just a hint.

REPLY [3 votes]: If $b\gt a$, then
$$
\begin{align}
\sum_{\substack{k\in\mathbb{Z}\\k\not\in\{-a,-b\}}}\frac1{(k+a)(k+b)}
&=\frac1{b-a}\lim_{n\to\infty}\sum_{\substack{k=-n\\k\not\in\{-a,-b\}}}^n\left(\frac1{k+a}-\frac1{k+b}\right)\\
&=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{\substack{k=-n+a\\k\not\in\{0,a-b\}}}^{n+a}\frac1k-\sum_{\substack{k=-n+b\\k\not\in\{0,b-a\}}}^{n+b}\frac1k\right)\\
&=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{\substack{k=-n+a\\k\ne0}}^{n+a}\frac1k-\sum_{\substack{k=-n+b\\k\ne0}}^{n+b}\frac1k+\frac2{b-a}\right)\\
&=\frac1{b-a}\lim_{n\to\infty}\left(\sum_{k=-n+a}^{-n+b-1}\frac1k-\sum_{k=n+a-1}^{n+b}\frac1k+\frac2{b-a}\right)\\[9pt]
&=\frac2{(b-a)^2}
\end{align}
$$
Switching $a$ and $b$ gives the same result for $b\lt a$.
If $a=b$, then
$$
\begin{align}
\sum_{\substack{k\in\mathbb{Z}\\k\ne-a}}\frac1{(k+a)^2}
&=\sum_{\substack{k\in\mathbb{Z}\\k\ne0}}\frac1{k^2}\\
&=2\sum_{k=1}^\infty\frac1{k^2}\\[6pt]
&=\frac{\pi^2}{3}
\end{align}
$$