Evaluation of double summation I am trying to evaluate the following summation : $\sum_{m=2}^{\infty}\sum_{n=1}^{m-1}\frac{1}{(2m-1)(2n-1)(2m-2n)}$
The above summation is a partial sum of another double summation, $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{1}{mn(m-n)}$. This second summation is equal to 2$\zeta(3)$. Hence, the first summation is convergent.
If we manipulate the summation that is in question, it turns out to be equal to $\sum_{m=1}^{\infty}\frac{H_{2m-1}}{(2m+1)^2}$ where $H_k$ represents the $k^{th}$ Harmonic number.

Here an alternative way:
$$\sum_{m\geq 2}\sum_{n=1}^{m-1}\frac{1}{(2m-1)(2n-1)(2m-2n)}=\frac{1}{2}\sum_{T\geq 1}\sum_{S=0}^{T-1}\frac{1}{(2T+1)(2S+1)(T-S)} $$
equals
$$ \frac{1}{2}\sum_{S\geq 0}\sum_{T>S}\frac{1}{(2T+1)(2S+1)(T-S)}=\frac{1}{2}\sum_{S\geq 0}\sum_{d\geq 1}\frac{1}{(2S+2d+1)(2S+1)d} $$
or
$$ \frac{1}{2}\sum_{S\geq 0}\sum_{d\geq 1}\int_{0}^{1}\frac{x^{2S}}{2S+1}\cdot\frac{x^{2d}}{d}\,dx=\frac{1}{2}\int_{0}^{1}\frac{\text{arctanh}(x)}{x}\cdot\left(-\log(1-x^2)\right)\,dx $$
or
$$ \frac{1}{4}\int_{0}^{1}\frac{\log^2(1+x)-\log^2(1-x)}{x}\,dx $$
where
$$ \int_{0}^{1}\frac{\log^2(1-x)}{x}\,dx = \int_{0}^{1}\frac{\log^2(x)}{1-x}\,dx = 2\zeta(3) $$
and
$$ \int_{0}^{1}\frac{\log^2(1+x)}{x}\,dx\stackrel{\text{IBP}}{=}-2\int_{0}^{1}\frac{\log(1+x)\log(x)}{1+x}\,dx  $$
is easily solved as $\frac{1}{4}\zeta(3)$ via $\log(1+x)=\log(1-x)-2\text{arctanh}(x)$, $x\mapsto\frac{1-x}{1+x}$ and
$$ \int_{0}^{1}\frac{\text{arctanh}(x)^k}{x}\,dx =\int_{0}^{+\infty}\frac{2t^k}{\sinh(2t)}\,dt=\frac{2k!}{2^k}\sum_{n\geq 0}\frac{1}{(2n+1)^{k+1}}=\frac{2k!}{2^k}\zeta(k+1)\left(1-\frac{1}{2^{k+1}}\right).$$
