Series

Let $a>\frac{1}{4}$. Prove that:

$$\sum_{n=1}^{\infty} \frac{1}{n^2-n+a} = \frac{\pi}{\sqrt{4a-1}} \frac{e^{\pi \sqrt{4a-1}}-1}{e^{\pi \sqrt{4a-1}}+1}$$

Solution

We note first of all that the roots of the denominator are $z_1=\frac{1+i \sqrt{4a-1}}{2}$ and $z_2=\frac{1-i \sqrt{4a-1}}{2}$. Now we easily see that $z_2=1-z_1$. Hence:

\begin{align*}
\sum_{n=1}^{N} \frac{1}{n^2-n+a} &=\frac{1}{z_1-z_2} \sum_{n=1}^{N} \left ( \frac{1}{n-z_1} - \frac{1}{n-z_2} \right ) \\
 &= \frac{1}{i\sqrt{4a-1}} \sum_{n=1}^{N} \left ( \frac{1}{n-z_1} + \frac{1}{z_2-n} \right ) \\
 &=\frac{1}{i\sqrt{4a-1}} \sum_{n=1}^{N} \left ( \frac{1}{n-z_1} + \frac{1}{1-z_2-n} \right ) \\
 &= \frac{1}{i \sqrt{4a-1}} \left ( \sum_{n=1}^{N} \frac{1}{n-z_1} + \sum_{n=-N+1}^{0} \frac{1}{n-z_1} \right )\\
 &= \frac{1}{i\sqrt{4a-1}} \sum_{n=-N+1}^{N} \frac{1}{n-z_1}
\end{align*}

Thus

\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{n^2-n+a} &= \lim_{N \rightarrow +\infty} \frac{1}{i\sqrt{4a-1}} \sum_{n=-N+1}^{N} \frac{1}{n-z_1} \\
 &= \frac{1}{i\sqrt{4a-1}} \sum_{n=-\infty}^{\infty} \frac{1}{n-z_1}\\
 &=\frac{1}{i\sqrt{4a-1}} \frac{\pi}{\tan \pi z_1} \\
 &=\frac{\pi}{\sqrt{4a-1}} \frac{e^{\pi \sqrt{4a-1}}-1}{e^{\pi \sqrt{4a-1}}+1}
\end{align*}