Calculate $\sum_{k=1}^\infty \frac{1}{k^2-L^2}$ and $\sum_{k=1}^\infty \frac{1}{\left(k-\frac{1}{2}\right)^2-L^2}$ Calculate
$$\sum_{k=1}^\infty \frac{1}{k^2-L^2}, \ \ \ \sum_{k=1}^\infty \frac{1}{\left(k-\frac{1}{2}\right)^2-L^2}$$
for $L<1/4$. 
The two series is always positive by $L<1/4$ and they obviously are converging. The problem, for me, is to calculate their sums. Suggestions are welcome.
 A: Hint:
$$\sum_{k=1}^{\infty }\frac{1}{k^2-L^2}=\frac{1}{L^2}\sum_{k=1}^{\infty }\frac{1}{(\frac{k^2}{L^2})-1}$$
$$\frac{\pi L}{2}\cot \pi L=\frac{1}{2}-\sum_{k=1}^{\infty }\frac{1}{k^2/L^2-1}$$
Laurent series see http://mathworld.wolfram.com/Cotangent.html
A: For the first sum, see e.g. the answers to this question (set $a=0$, $b=iL$). The result is 
$$S_1(L)=\sum_{k=1}^{\infty}\frac{1}{k^2-L^2}=\frac12\left(\frac1{L^2}+\sum_{k=-\infty}^{\infty}\frac{1}{k^2-L^2}\right)=
\frac12\left(\frac1{L^2}-\frac{\pi\cot\pi L}{L}\right).
$$
The second sum can be expressed in terms of the first one. Indeed, 
\begin{align*}
S_2(L)&=\sum_{k=1}^{\infty}\frac{4}{(2k-1)^2-4L^2}=
4\sum_{k=1}^{\infty}\frac{1}{k^2-4L^2}-4\sum_{k=1}^{\infty}\frac{1}{(2k)^2-4L^2}=\\&=4S_1(2L)-S_1(L)=-\frac{\pi\cot 2\pi L}{L}+\frac{\pi\cot \pi L}{2L}=\frac{\pi\tan\pi L}{2L}.
\end{align*}
A: Hint: First read the article on the Basel problem in order to understand the basic intuition behind Euler's infinite product formula for the sine function. Then try so write a similar product for the cosine function as well, using the same ingenious approach. Afterwards, differentiate the natural logarithm of each.
