Infinite sums of squares $$\sum_{n=0}^{\infty} \frac {k^2(1-k)^2}{(n+k)^2(n+1-k)^2}$$
Here can anyone help me to solve this question,I can't think of any logic like telescopic, coefficient compare etc .
It would be helpful if anyone could provide the solution

The series diverges for integer $k$, Assuming $k \in \mathbb{R} \setminus \mathbb{Z}$, the series sums to $$\frac{k^2(1-k)^2}{(1-2k)^2}\left[\frac{\pi^2}{\sin^2(\pi k)} - \frac{2\pi\cot(\pi k)}{1-2k}\right]\tag{*1}$$
Let $\alpha = k$ and $\beta = 1 - k$, the sum at hand equals to
$$\begin{align}\verb/sum/ 
&= \alpha^2\beta^2\sum_{n=0}^\infty \frac{1}{(n+\alpha)^2(n+\beta)^2}\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)^2\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left(\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2} - \frac{2}{(n+\alpha)(n+\beta)}\right)\\
&= \frac{\alpha^2\beta^2}{(\alpha-\beta)^2}\sum_{n=0}^\infty\left[\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2} - \frac{2}{\beta - \alpha}\left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)\right]
\end{align}
$$
The four pieces inside above bracket can be combined into two pieces
$$\sum_{n=0}^\infty \left(\frac{1}{(n+\alpha)^2} + \frac{1}{(n+\beta)^2}\right)
= \sum_{n=0}^\infty \left(\frac{1}{(n+k)^2} + \frac{1}{(-(n+1)+k)^2}\right) = \sum_{n=-\infty}^\infty\frac{1}{(n+k)^2}$$
and
$$\sum_{n=0}^\infty \left(\frac{1}{n+\alpha} - \frac{1}{n+\beta}\right)
= \sum_{n=0}^\infty \left(\frac{1}{n+k} + \frac{1}{-(n+1)+k}\right) = \sum_{n=-\infty}^\infty\frac{1}{(n+k)^2}$$
where sums of the form $\sum\limits_{n=-\infty}^\infty (\cdots)$ is a short hand for the symmetric limit $\lim\limits_{p\to\infty}\sum\limits_{n=-p}^p (\cdots)$
Using following Mittag-Leffler expansion for cotangent and its derivative, $(*1)$ follows immediately.
$$\pi \cot(\pi x) = \sum_{n=-\infty}^\infty \frac{1}{x + n}
\quad\text{ and }\quad
\frac{\pi^2}{\sin^2(\pi x)} = \sum_{n=-\infty}^\infty \frac{1}{(x+n)^2}
$$
