The Value of a series

Question

What is the value of the following series

$\sum_{n=1}^\infty\sum_{m=1}^\infty\sum_{k=1}^\infty \frac{1}{mnk(m+n+k+1)}$

Answer 1

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: $$\sum _{m=1}^{\infty } \left( \sum _{k=1}^{\infty } \left( \sum _{n=1}^ {\infty }{\frac {1}{ \left( n+k+m+1 \right) nkm}} \right) \right) = \int _{0}^{\infty }\!{u}^{3}{{\rm e}^{-u}}{du}=\Gamma(4)=3!=6$$