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?

We show here how to solve the original problem.
Partial fraction decomposition gives
$$\frac{1}{k (k+1)}=\frac{1}{k}-\frac{1}{k+1}\tag{1}$$
Hence the sum can be written as
$$s = \sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right) H_{k+1}$$
$$=\sum _{k=1}^{\infty } \left(\frac{H_{k+1}}{k}-\frac{H_{k+1}}{k+1}\right)$$
Now we have
$$H_{k+1}=H_{k}+\frac{1}{k+1}$$
so that $s=s_{1} + s_{2}$ where
$$
\begin{align}
s_{1}&=\sum _{k=1}^{\infty } \left(\frac{H_k}{k}-\frac{H_{k+1}}{k+1}\right)\\
&=( \frac{H_1}{1}-\frac{H_2}{2})+(\frac{H_2}{2}-\frac{H_3}{3}) +\text{...}  =\frac{H_1}{1}=1\\
s_{2}&=\sum _{k=1}^{\infty } \frac{1}{k (k+1)}=\sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right)\\
& =( \frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}) +\text{...} =\frac{1}{1}=1 
\end{align}$$
As we can see both series telescope so that we get $s_{1}=1$ and $s_{2}=1$
so that finally $s=2$.
