How can we compute $\sum_{m=1}^\infty \frac{H_{\left(m-3/2\right)}}{m^2}$? In the evaluation of $$\sum_{k=1}^\infty \sum_{\ell=1}^{k-1}\sum_{m=1}^{\ell-1}\frac{\delta_{k, 2\ell-2m}}{m\left(\ell-m\right)\left(k-\ell\right)}.$$
Here $\delta_{k, 2\ell-2m}$ denotes the "Kronecker delta" (see here) and is given by 
$$\delta_{k, 2\ell-2m} = \begin{cases}
0 \qquad \mathrm{if} k \neq 2\ell-2m, \\
1 \qquad \mathrm{if} k =2\ell-2m.
\end{cases}$$
Thus we have the equality 
$$S = \sum_{\ell=1}^\infty\sum_{m=1}^{\ell-1} \frac{1}{m(\ell-m)(\ell-2m)}.$$
I was able to simplify this down to something along the lines of
$$\sum_{n=1}^\infty \frac{H_{\left(n-3/2\right)}}{n^2}$$which I have no idea how to approach. Sorry for the little information given, I am just looking for an idea of how I can go about this. Thank you!
 A: You did not reduce the sum properly. For a given even $k$, the set of $m$ that satisfies $k = 2\ell - 2m, 1 \le \ell \le k-1, 1 \le m \le \ell-1$ is the positive natural numbers less than or equal to $k-1$
This means that the sum should be $$\sum_{t = 1}^{\infty} \sum_{m=1}^{t-1} \frac{1}{m(\ell-m)(k-\ell)}$$
where $k = 2t$ and $\ell = \frac{k+2m}{2} = t+m$
Plugging those in and simplifying, I get $$\sum_{t = 1}^{\infty} \sum_{m=1}^{t-1} \frac{1}{mt(t-m)}$$
Changing the order of summation yields $$\sum_{m=1}^{\infty} \sum_{t=m+1}^{\infty} \frac{1}{mt(t-m)}$$
The inner sum can be simplified so that the double sum becomes $$\sum_{m=1}^{\infty}\frac{H_m}{m^2}$$
Using this answer to another question, the sum can be simplified to $$2\zeta(3) \approx 2.404$$
A: It is not a complete answer. Let
\begin{eqnarray}
f(x)&=&\sum_{m=1}^\infty \frac{H_{m-\frac32}}{m^2}x^m
\end{eqnarray}
and then
\begin{eqnarray}
f'(x)&=&\sum_{m=1}^\infty \frac{H_{m-\frac32}}{m}x^{m-1}\\
(xf'(x))'&=&\sum_{m=1}^\infty H_{m-\frac32}x^{m-1}=\sum_{m=1}^\infty x^{m-1}\int_0^1\frac{1-t^{m-\frac32}}{1-t}dt\\
&=&\int_0^1\sum_{m=1}^\infty \frac{1-t^{m-\frac32}}{1-t}x^{m-1}dt\\
&=&\int_0^1\bigg(\frac{1}{(1-t) (1-x)}-\frac{1}{(1-t)\sqrt t (1-t x)}\bigg)dt.
\end{eqnarray}
So
\begin{eqnarray}
xf'(x)&=&\int_0^x\int_0^1\bigg(\frac{1}{(1-t) (1-r)}-\frac{1}{(1-t) \sqrt t(1-t r)}\bigg)dtdr\\
&=&\int_0^1\int_0^x\bigg(\frac{1}{(1-t) (1-r)}-\frac{1}{(1-t)\sqrt t (1-t r)}\bigg)drdt\\
&=&\int_0^1\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dt\\
f(1)&=&\int_0^1\frac1x\int_0^1\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dtdx\\
&=&\int_0^1\int_0^1\frac1x\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dxdt\\
&=&\int_0^1\frac{\pi ^2-6 t^{-3/2} \text{Li}_2(t)}{6-6 t}dt.
\end{eqnarray}
Now I have problem for the last step. But Wolfram Mathematica gives
$$ \int_0^1\frac{\pi ^2-6 t^{-3/2} \text{Li}_2(t)}{6-6 t}dt=\frac{7 \zeta (3)}{2}-\frac{1}{3} \pi ^2 (\log (2)-1)-8 \log (2). $$
