The Value of a series What is the value of the following series 
$\sum_{n=1}^\infty\sum_{m=1}^\infty\sum_{k=1}^\infty \frac{1}{mnk(m+n+k+1)}$
 A: Note that because the series for the logarithm is:
$$-\sum _{n=1}^{\infty }{\frac {{x}^{n}}{n}}=\ln  \left( 1-x \right) $$
it follows that:
$$\sum _{m=1}^{\infty } \left( \sum _{k=1}^{\infty } \left( \sum _{n=1}^
{\infty }{\frac {{x}^{n+k+m}}{nkm}} \right)  \right) =- \ln 
 \left( 1-x \right) ^{3}$$
integrating between $x=0$ and $x=1$ shows that:
$$\sum _{m=1}^{\infty } \left( \sum _{k=1}^{\infty } \left( \sum _{n=1}^
{\infty }{\frac {1}{ \left( n+k+m+1 \right) nkm}} \right)  \right) =
\int _{0}^{1}\!-  \ln  \left( 1-x \right) ^{3}{dx}$$
the integral on the right can be evaluated, one way is to make the substitution: $x=-e^{-u}+1$ to get:
$$\int _{0}^{1}\!-  \ln  \left( 1-x \right) ^{3}{dx}=\int _{0}^{\infty }\!{u}^{3}{{\rm e}^{-u}}{du}$$
the integral on the right can be done quite simply using integration by parts or by recognising it as the $\Gamma$ function and we find:
$$\sum _{m=1}^{\infty } \left( \sum _{k=1}^{\infty } \left( \sum _{n=1}^
{\infty }{\frac {1}{ \left( n+k+m+1 \right) nkm}} \right)  \right) =
\int _{0}^{\infty }\!{u}^{3}{{\rm e}^{-u}}{du}=\Gamma(4)=3!=6$$
