<s>[INST] How find this sum $\sum_{n=1}^{\infty}\frac{1}{n^2-n+a}$ Today Question
if $a>\dfrac{1}{4}$, show that
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2-n+a}=\dfrac{\pi}{\sqrt{4a-1}}\cdot\dfrac{e^{\pi\sqrt{4a-1}}-1}{e^{\pi\sqrt{4a-1}}+1}\tag{1}$$
I have  konw that solution:

$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$

solution:
note

$$e^{ax}=\dfrac{\pi^{2\pi a}-1}{\pi}\left(\dfrac{1}{2a}+\sum_{n=1}^{\infty}\dfrac{a\cos{nx}-n\sin{x}}{n^2+a^2}\right)$$
  let $a=1,x=0$,then we have
  $$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$
But for $(1)$ I have
  $$\sum_{n=1}^{\infty}\dfrac{1}{(n-\frac{1}{2})^2+a-\frac{1}{4}}$$
  Then I can't ,Thank you for your help.

 [/INST] Complete the square in the summand to get
$$\sum_{n=1}^{\infty} \frac{1}{\left ( n-\frac12\right)^2+\left (a-\frac14\right)}$$
Note that the sum is symmetric and equal to
$$\frac12 \sum_{n=-\infty}^{\infty} \frac{1}{\left ( n-\frac12\right)^2+\left (a-\frac14\right)}$$
so that we may use the residue theorem to evaluate the sum.  I will skip the steps involved in deriving the general result, but will just state it here:
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{k} \operatorname*{Res}_{z=z_k} \pi \, \cot{(\pi z)} \, f(z)$$
where the $z_k$ are poles of $f$, which obeys certain integrability conditions.  Details may be found elsewhere within this site.
In this case,
$$f(z) = \frac{1}{\left ( z-\frac12\right)^2+\left (a-\frac14\right)}$$
which has poles $z_{\pm} = \frac12 \pm i \sqrt{a-\frac14}$.  The sum is therefore equal to
$$-\frac{\pi}{2} 2 \Re{\left [\frac{\cot{\pi \left ( \frac12 + i \sqrt{a-\frac14}\right )}}{i 2 \sqrt{a-\frac14}} \right ]} = \frac{\pi}{\sqrt{4 a-1}} \tanh{\left ( \pi \sqrt{4 a-1}\right)}$$
which matches the result to be shown.
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