Expressing Tornheim sums in terms of Riemann's Zeta If 
$$T(a,b,c)=\sum_{r\geq1}\sum_{s\geq1} \frac{1}{r^as^b(r+s)^c}$$
How to prove that :
$$T(3,1,2)=-\frac13 \zeta(6)+\frac{\zeta^2(3)}{2}$$
I tried some algebraic manipulations but did not work. Can you please help me ?
Any solution will be appreciated 
 A: You might benefit from reading Basu, A., "On the evaluation of Tornheim sums and allied double sums", doi:10.1007/s11139-011-9302-5.  Table 2, item F would be a starting point.
A: The easiest way to do it is to use partial fraction decomposition. Let us choose the variable for the decomposition to be $s$. Then we have:
\begin{equation}
\frac{1}{r^a s^b (s+r)^c} = 
\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} \frac{(-1)^{b-l_1}}{s^{l_1} r^{b+b+c-l_1}} +
\sum\limits_{l_1-1}^c \binom{b+c-1-l_1}{b-1} \frac{(-1)^b}{(s+r)^{l_1} r^{a+b+c-l_1}}
\end{equation}
Now we sum over $s$ and we get :
\begin{equation}
\sum\limits_{s=1}^\infty \frac{1}{r^a s^b (s+r)^c} = 
\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} \frac{(-1)^{b-l_1} \zeta(l_1) 1_{l_1 \ge 2}}{ r^{a+b+c-l_1}} +
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} \frac{(-1)^b(\zeta(l_1) 1_{l_1 \ge 2} - H_r^{(l_1)})}{ r^{a+b+c-l_1}}
\end{equation}
I think that this that doesn't require any explanation except that for $l_1=1$ we are getting singular terms. Therefore in that case we simply apply the following identity $H_r^{(1)} = \sum_{s\ge 1}\left( 1/s - 1/(s+r)\right)$. Now we sum over the variable $r$ and we get:
\begin{eqnarray}
&&T(a,b,c)=\sum\limits_{l_1=1}^b \binom{b+c-1-l_1}{c-1} (-1)^{b-l_1} \zeta(l_1) 1_{l_1 \ge 2} \zeta(a+b+c-l_1) +
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} (-1)^b\zeta(l_1) 1_{l_1 \ge 2}  \zeta(a+b+c-l_1) - \\
&&
\sum\limits_{l_1=1}^c \binom{b+c-1-l_1}{b-1} (-1)^b {\bf H}^{(l_1)}_{a+b+c-l_1}(+1)
\end{eqnarray}
where ${\bf H}^{(l)}_n(t) := \sum\limits_{r\ge 1} H_r^{(l)}/r^n t^r$.
Now if we take $(a,b,c)=(3,1,2)$ we get:
\begin{eqnarray}
T(3,1,2)&=& 0 + (-1)^1 \zeta(2) \zeta(4) - \sum\limits_{l_1=1}^2 \binom{2-l_1}{0} (-1)^1 {\bf H}^{(l_1)}_{6-l_1}(+1)\\
&=&(-1)^1 \zeta(2) \zeta(4)+ {\bf H}^{(1)}_5(+1) + {\bf H}^{(2)}_4(+1) \\
&=&(-1)^1 \zeta(2) \zeta(4)+\left(-\frac{1}{2} \zeta(3)^2-1/3 \zeta(2) \zeta(4)+\frac{7}{3} \zeta(6)\right)+\left(+1 \zeta(3)^2 + \frac{4}{3} \zeta(2) \zeta(4)-\frac{8}{3} \zeta(6)\right) \\
&=& \frac{1}{2} \zeta(3)^2 - \frac{1}{3} \zeta(6)
\end{eqnarray}
where in the second line from the bottom we used my answer to Calculating alternating Euler sums of odd powers .
