<s>[INST] Find triple summation rel in a closed form $S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$ 
Evaluate $\displaystyle S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$ 

My attempt : 
Let  $$A=\sum_{k=1}^{m}\frac{1}{k(k+1)} 
=\sum_{k=1}^{m}\left( \frac1{k}-\frac1{k+1} \right) = \frac{m}{m+1}$$
and a second sum : 
$$B=\sum_{m=1}^{n}\frac{1}{(m+1)^{2}}$$
from here how I can complete ??
 [/INST] Continuing from the calculation above we have to sum:
$$S=\sum_{n=1}^\infty\frac{1}{n(n+1)}\sum_{m=1}^n\frac{1}{(m+1)^2}=\sum_{n=1}^{\infty}\frac{H_2(n+1)-1}{n(n+1)}\\=\sum_{n=1}^{\infty}[\frac{H_2(n+1)}{n}-\frac{H_2(n+1)}{n+1}]-\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$
Using the identity $H_2(n+1)=H_2(n)+\frac{1}{(n+1)^2}$ 
$$S=\sum_{n=1}^{\infty}[\frac{H_2(n)}{n}-\frac{H_2(n+1)}{n+1}]-1+\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}$$
And since the first sum telescopes to $1$ and the identity $\frac{1}{n(n+1)^2}=\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}$ holds we obtain that:
$$S=\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}=\sum_{n=1}^{\infty}\Big[\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}\Big]=1-(\zeta(2)-1)=2-\frac{\pi^2}{6}$$
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