"Compute $$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}}$$ using contour integration"
I have used the function $F(z) = \frac {\pi cot\pi z}{z^2(z+1)^2}$
Which has double poles at $z=0$ and $z=-1$
For the pole at $z=0$, if I calculate the residue by taking the limit of $\frac{dF}{dz}$ as $z \to 0 $, I end up with a $cosec(0)$ term, which is $\infty$
Instead I can try to calculate the residue using the Laurent series, and about $z=0$ again I find;
$\frac {\pi cot(\pi z)}{z^2{(z+1)^2}} = \frac{\pi}{z^2} [(\pi z)^{-1} - \frac {1}{3}(\pi z) - \frac {1}{45}(\pi z)^3 ...][1 - 2z +3z^2 -...]$
And I find the residue to be $3-\frac{1}{3} \pi^2$
To compute the residue at $z=-1$, however, I can't compute the expansion of $\pi cot(\pi z)$ about $z=-1$ using normal Taylor expansion methods, because if:
$g(z) = \pi cot(\pi z)$, then $g(-1) = \frac{1}{0}$
and that's where I'm stuck - computing the residue at $z=-1$.
Once I've found the residue, computing the series is simple:
$$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}} = \frac{1}{2} \sum_{Res}$$
Any help would be greatly appreciated!