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
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.
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 .
