<s>[INST] Is there any way to simplify following summation? Is there any way to simplify following summation?
$$\sum_{k=1}^n \frac{1}{k^2(k+1)^2}$$
 [/INST] Consider 
\begin{align}
\frac{1}{k^2 (k+1)^2} = \left(\frac{1}{k} - \frac{1}{k+1}\right)^{2} = \frac{1}{k^2} + \frac{1}{(k+1)^2} - \frac{2}{k} + \frac{2}{k+1}
\end{align}
for which
\begin{align}
\sum_{k=1}^{n} \frac{1}{k^2 (k+1)^2} &= \sum_{k=1}^{n} \frac{1}{k^2} + \sum_{k=2}^{n+1} \frac{1}{k^2} + 2 \sum_{k=2}^{n+1} \frac{1}{k} - 2 \sum_{k=1}^{n} \frac{1}{k} \\
&= -1 - \frac{1}{(n+1)^2} + 2 \sum_{k=1}^{n+1} \frac{1}{k^2} + \frac{2}{n+1} - 2 \\
&= -2 - \left( \frac{1}{n+1} - 1\right)^2 + \sum_{k=1}^{n+1} \frac{2}{k^2}\\
&= - \frac{3n^2 + 4n +2}{(n+1)^2} + \sum_{k=1}^{n+1} \frac{2}{k^2} \\
&= \frac{\pi^2}{6} - \frac{3n^2 + 4n +2}{(n+1)^2} - \psi_{1}(n+2)
\end{align}
where $\psi_{1}(x)$ is the derivative of the digamma function. 
</s>