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
 A: 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.
$$\pi \cot(\pi x) = \sum_{n=-\infty}^\infty \frac{1}{x + n}
\quad\text{ and }\quad
\frac{\pi^2}{\sin^2(\pi x)} = \sum_{n=-\infty}^\infty \frac{1}{(x+n)^2}
$$
A: Mathematica gives the following result:
$$\sum\limits_{n=0}^{\infty}\frac{k^2 (1-k)^2}{(n+k)^2 (n+1-k)^2}=\frac{k^2 (1-k)^2 \left(\pi \csc^2(\pi k) (\pi (2 k-1)+\sin(2 \pi k))\right)}{(2 k-1)^3}\ ,$$ $$\Im(k)\neq 0\lor 0<k<1\lor k\notin \mathbb{Z}$$
A: I thought it would be instructive to present a way forward using contour integration.  To that end we now proceed.

Let $f(z)=\frac{\cot(\pi z)}{(z+k)^2(z+1-k)^2}$ and let $N\ge \max(|k|, |k-1|)$.  Then, we see that
$$\begin{align}\oint_{|z|=N+1/2}f(z)\,dz&=2\pi i \left(\text{Res}\left(f(z), z=-k\right)+\text{Res}\left(f(z), z=k-1\right)+\sum_{n=-\infty}^\infty \text{Res}\left(f(z), z=n\right)\right)\\\\
&=2\pi i \left(\text{Res}\left(f(z), z=-k\right)+\text{Res}\left(f(z), z=k-1\right)+ \sum_{n=-\infty}^\infty \frac{1}{\pi(n+k)^2(n+1-k)^2}\right)\\\\
&=2\pi i \left(\text{Res}\left(f(z), z=-k\right)+\text{Res}\left(f(z), z=k-1\right)+ \sum_{n=0}^\infty \frac{2}{\pi(n+k)^2(n+1-k)^2}\right)\tag1
\end{align}$$
Note that $|\cot(\pi z)|$ is bounded on the circle $|z|=N+1/2$.  Then, letting $N\to \infty$ in $(1)$, we find that
$$\sum_{n=0}^\infty \frac{1}{(n+k)^2(n+1-k)^2}=\frac{\pi}{2}\left(-\text{Res}\left(f(z), z=-k\right)-\text{Res}\left(f(z), z=k-1\right)\right)\tag2$$

The residues in $(2)$ are given by
$$\begin{align}
\text{Res}\left(f(z), z=-k\right)&=\lim_{z\to -k}\frac{d}{dz}\frac{\cot(\pi z)}{(z+1-k)^2}\\\\
&=\frac{-\pi \csc^2(\pi k)}{(2k-1)^2}-2\frac{\cot(k\pi)}{(2k-1)^3}\tag{3a}\\\\
\text{Res}\left(f(z), z=k-1\right)&=\frac{-\pi \csc^2(\pi k)}{(2k-1)^2}+2\frac{\cot(k\pi)}{(2k-1)^3}\tag{3b}
\end{align}$$

Using $(3a)$ and $(3b)$ reveals that
$$\sum_{n=0}^\infty \frac1{(n+k)^2(n+1-k)^2}=\frac{\pi^2 \csc^2(\pi k)}{(2k-1)^2}-\frac{2\pi \cot(\pi k)}{(2k-1)^3}\tag4$$
whereupon multuplying $(4)$ by $k^2(1-k)^2)$ yields the coveted result.
A: Assuming $k$ is an integer:

*

*From the $n+k$ in the denominator it follows that $k \geqslant 1$ because otherwise one summand is undefined.


*From the $n+1-k$ in the denominator it follows that $1-k \geqslant 1$ because otherwise one summand is undefined.
Not much that's left for $k$...
A: (I don't see an equation, just an expression?) Convert the expression into partial fractions
$
...= \frac{2k^2(1-2k+k^2)}{(-1+2k)^3} \frac{1}{n+k}-\frac{2k^2(1-2k+k^2)}{(-1+2k)^3} \frac{1}{n+1-k}+\frac{k^2(1-2k+k^2)}{(-1+2k)^2}\frac{1}{(n+k)^2}+
\frac{k^2(1-2k+k^2)}{(-1+2k)^2}\frac{1}{(n+1-k)^2}
$
and the result can be written as a sum over $\psi(k)$,
$\psi(1-k)$ and its derivatives.
