Proving that $\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$ I'm trying to prove that:
$$\frac{3}{2} \sum_{k=1}^{\infty} \frac{4}{k^3+k^2} = \pi^2-6$$
I've tried looking at the partial sums, but no luck there. I just have no idea where to begin.
Knowing that $\frac{4}{k^3+k^2} = \frac{4}{k^2 (k+1)} \implies \frac{4}{k^2}+\frac{4}{(k+1)}-\frac{4}{k}$ seemed to help, but I get stuck no matter how I look at it.
Any help would be greatly appreciated.
 A: Hint:
$$
    \frac{4}{k^3+k^2} = \frac{4}{k^2(k+1)} = \frac{a_1}{k^2} + \frac{a_2}{\vphantom{k^2}k} + \frac{a_3}{\vphantom{k^2}k+1}
$$
for some constants $a_1$, $a_2$, and $a_3$, and
$$
    \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}.
$$
A: You almost got it. 
The sum $\sum\limits_{k=1}^{\infty} \frac 4 {k+1}-\frac 4 k$ is a telecscopic sum, so its easy.
for the sum $\sum\limits_{k=1}^{\infty} \frac 4 {k^2}$ you need an expansion in fourier series and to apply Parseval's theorem and you are done. Try expanding $f(x)=x$ in $(-\pi, \pi)$, I'm not sure if that interval is the best, but it should do the trick.
Also $\sum\limits_{k=1}^{\infty} \frac 1 {k^2} = \zeta(2)=\frac {\pi ^2} 6$ 
A: Equivalently, you are asking for $$6\,\sum_{k=1}^{\infty}\frac{1}{k^3 + k^2} $$
Using your partial fractions decomposition, we can split this into $$6\,\sum_{k=1}^{\infty}\left(\frac{1}{k^2} + \frac{1}{k + 1} - \frac{1}{k}\right) $$
The second and third terms telescope, giving a sum of -1. The first term sums to $\pi^2/6$, which is a well-known fact. For a number of proofs, see here.
Putting this all together, your series sums to $\pi^2 - 6$. You can complete this argument.
