Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $

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 $$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}=1.$$

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 $s=2$.