<s>[INST] Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ First, the original problem follows, 
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$
where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series.
Using the following identity,
$$H_{k+1} = \sum_{j=1}^\infty (\frac{1}{j} - \frac{1}{k+j+1}).$$
I was able to get this one.
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}=\sum_{j=1}^\infty [\frac{1}{j} - \frac{1}{j+1} + \frac{1}{j+1}\sum_{k=1}^\infty \frac{1}{(k+1)(k+1+j)}] $$
So, If I get the sum of the double series, 
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}.$$
I can also find the original problem.
What method can I use at this problem?
 [/INST] Going back to the original problem, note that for any positive integer $N$,
$$
\begin{align}
\sum_{k=1}^N \frac{H_{k+1}}{k(k+1)}&=\sum_{k=1}^N \frac{H_{k+1}}{k}-\sum_{k=1}^N \frac{H_{k+1}}{k+1}\\
&=\sum_{k=1}^N \frac{H_{k}}{k}+\sum_{k=1}^N \frac{1}{k(k+1)}-\sum_{k=2}^{N+1} \frac{H_{k}}{k}\\
&=1+\sum_{k=1}^N \left(\frac{1}{k}-\frac{1}{k+1}\right)- \frac{H_{N+1}}{N+1}\\
&=2-\frac{1}{N+1}- \frac{H_{N+1}}{N+1}.
\end{align}$$
Hence, by taking the limit as $N\to +\infty$, we get
$$\sum_{k=1}^{\infty} \frac{H_{k+1}}{k(k+1)}=2.$$
P.S. It follows that 
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}=1.$$
</s>