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?
 A: 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 
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}=1.$$
A: 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 $s=2$.
A: As an alternative approach,
$$S=\sum_{j,k\geq 1}\frac{1}{(j+1)(k+1)(j+k+1)}=\sum_{j,k\geq 1}\iiint_{(0,1)^3}x^j y^k z^{j+k}\,dx\,dy\,dz$$
and Fubini's theorem allows to switch the double series and the triple integral (this sounds a bit ludicrous, I know), leading to
$$\begin{eqnarray*} S &=&\iiint_{(0,1)^3}\frac{xyz^2}{(1-xz)(1-yz)}\,dx\,dy\,dz\\&=&\int_{0}^{1}\iint_{(0,z)^2}\frac{XY}{(1-X)(1-Y)}\,dX\,dY\,\frac{dz}{z^2}\\ 
&=&\int_{0}^{1}\left[\int_{0}^{z}\frac{w}{1-w}\,dw\right]^2\,\frac{dz}{z^2} \\&=&\int_{0}^{1}\left[1+\frac{\log(1-z)}{z}\right]^2\,dz\\&=&1-2\,\zeta(2)+\int_{0}^{1}\frac{\log^2(1-z)}{z^2}\,dz\\&=&1-2\,\zeta(2)+\int_{0}^{1}\frac{\log^2(z)}{(1-z)^2}\,dz\\&=&1-2\,\zeta(2)+\sum_{m\geq 1}\int_{0}^{1}m z^{m-1}\log^2(z)\,dz\\&=&1-2\,\zeta(2)+2\sum_{m\geq 1}\frac{1}{m^2}=\color{red}{\large 1}.\end{eqnarray*}$$
A: $$\frac{H_{k+1}}{k(k+1)}=\frac{H_{k+1}}k-\frac{H_{k+1}}{k+1}=\frac{H_k}k+\frac1k-\frac1{k+1}-\frac{H_{k+1}}{k+1}$$ nicely telescopes to the limit
$$\frac{H_1}1+\frac11.$$
