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.

Here's another approach. 
It depends primarily on the properties of telescoping series, partial fraction expansion, and the following identity for the $m$th harmonic number
$$\begin{eqnarray*}
\sum_{k=1}^\infty \frac{1}{k(k+m)}
&=& \frac{1}{m}\sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+m}\right) \\
&=& \frac{1}{m}\sum_{k=1}^m \frac{1}{k} \\
&=& \frac{H_m}{m},
\end{eqnarray*}$$
where $m=1,2,\ldots$.
Then,
$$\begin{eqnarray*}
\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}
&=& \sum_{k=1}^{\infty} \frac{1}{k}
    \sum_{n=1}^{\infty} \frac{1}{n(n+k+2)} \\
&=& \sum_{k=1}^{\infty} \frac{1}{k} \frac{H_{k+2}}{k+2} \\
&=& \frac{1}{2} \sum_{k=1}^{\infty} 
    \left( \frac{H_{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\
&=& \frac{1}{2} \sum_{k=1}^{\infty}
    \left( \frac{H_k +\frac{1}{k+1}+\frac{1}{k+2}}{k} - \frac{H_{k+2}}{k+2} \right) \\
&=& \frac{1}{2} \sum_{k=1}^{\infty}
    \left( \frac{H_k}{k} - \frac{H_{k+2}}{k+2} \right)
    + \frac{1}{2} \sum_{k=1}^{\infty}
    \left(\frac{1}{k(k+1)} + \frac{1}{k(k+2)}\right) \\
&=& \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) 
    + \frac{1}{2}\left(H_1 + \frac{H_2}{2}\right) \\
&=& \frac{7}{4}. 
\end{eqnarray*}$$
