Compute: $\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$ I try to solve the following sum:
$$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$
I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.

I think one way of approaching this sum would be to use the partial fraction 
$$ \frac{1}{k^2n+2nk+n^2k} = \frac{1}{kn(k+n+2)} =  \frac{1}{2}\Big(\frac{1}{k} + \frac{1}{n}\Big)\Big(\frac{1}{k+n} - \frac{1}{n+k+2}\Big)$$
to rewrite you sum in the form 
$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}  \frac{1}{k^2n+n^2k+2kn} = \frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \Big( \frac{1}{n(k+n)} - \frac{1}{n(k+n+2)} + \frac{1}{k(k+n)} - \frac{1}{k(k+n+2)}\Big)$$
Since the sum on the right will telescope in one of the summation variables it should be straightforward to find the answer from here (it ends up being $7/4$ I think).
