<s>[INST] How can one evaluate the basic Tornheim Sum? I recently came across the claim that the double sum from $1$ to $\infty$ of $\frac{1}{mn(m+n)} = 2 \zeta(3)$.
I can show it equals the sum from $1$ to $\infty$ of $\frac{H_n}{n^2}$, where $H_n$ is the $n^{th}$ Harmonic number.
Does anyone know the full proof?
 [/INST] Consider the following three sums 
\begin{eqnarray*}
T(1,1,1) & =&  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)} \\
\zeta(2,1) & =&  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2} \\
\zeta(3) & =&  \sum_{m=1}^{\infty}  \frac{1}{m^3}. \\
\end{eqnarray*}
Using partial fractions, you already know
\begin{eqnarray*}
  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)}  =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2} \left(\frac{1}{n} -\frac{1}{m+n} \right)=  \sum_{m=1}^{\infty}  \frac{H_m}{m^2} = \zeta(3) + \zeta(2,1) .
\end{eqnarray*}
Now the trick you need is this
\begin{eqnarray*}
  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)}  =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \color{red}{\frac{1}{(m+n)^2} \left(\frac{1}{m} +\frac{1}{n} \right) } \\ =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2}  +\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n(m+n)^2}  = 2 \zeta(2,1) 
\end{eqnarray*}
So $\zeta(2,1)=\zeta(3)$ & your sum is $T(1,1,1) =2 \zeta(3)$.
</s>