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 ??

Just for your curiosity.
We could also have a cloased form for the partial sum
$$S_p=\sum_{n=1}^{p}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$$since
$$\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}=\frac{1}{(m+1)^2 n (n+1)}$$
$$\sum_{m=1}^{n}\frac{1}{(m+1)^2 n (n+1)}=\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}$$
$$S_p=\sum_{n=1}^{p}\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}=\frac{12 (p+1)-\pi ^2 (p+2)+6 (p+2) \psi ^{(1)}(p+2)}{6 (p+1)}$$
Now, using, for large values of $q$, the expansion
$$\psi ^{(1)}(q)=\frac{1}{q}+\frac{1}{2 q^2}+\frac{1}{6 q^3}+O\left(\frac{1}{q^5}\right)$$ and continuing with Taylor series for large values of $p$
$$S_p=\left(2-\frac{\pi ^2}{6}\right)+\frac{1-\frac{\pi ^2}{6}}{p}+\frac{\pi ^2-3}{6
   p^2}-\frac{2+\pi ^2}{6 p^3}+\frac{10+\pi ^2}{6 p^4}+O\left(\frac{1}{p^5}\right)$$
For illustration purposes, using $p=10$, the exact value is
$$S_{10}=\frac{42308191}{140873040}\approx 0.300328$$ while the above truncated expansion gives
$$\frac{125690-10909 \pi ^2}{60000}\approx    0.300375$$
