Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that 


  $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$


This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.



Any way, I tried hard on this one but no success yet. here is what I did:

Using the identity  

$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$

$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$

We can write 

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$

where

\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&amp;=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&amp;=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&amp;=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&amp;=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}

Therefore 

$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$

The sum can be done using the following identity

$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a &gt;0.$$

Differentiate both sides with respect to $a$ then set $a=1/2$ we get 

$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$



and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks
Cornel's way to make it easy. Replace the harmonic number in the numerator by Digamma function, using that $H_{n/2}= \psi(n/2+1)+\gamma$, and then splitting the series using the parity, we have 

$$ S=\sum_{n=1}^{\infty} \frac{ \psi(n/2+1)+\gamma}{(2n+1)^3}=\sum_{n=1}^{\infty} \frac{ \psi(n+1)+\gamma}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ \psi(n+1/2)+\gamma}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ 2H_{2n}-H_n-2\log(2)}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}-\sum_{n=1}^{\infty} \frac{H_n}{(4n-1)^3}-2\log(2)\sum_{n=1}^{\infty} \frac{1}{(4n-1)^3}+2\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3},$$
and since the first two series are straightforward using Cornel's Master Theorem of Series from A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series, and then noting that 

$$\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3}=\frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{H_{n}}{(2n-1)^3}-\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n-1)^3}\right),$$

where for the first series we can use the same mentioned master theorem, and then the second one is already known in the form  $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n+1)^3}$ (it's easy to rearrange the series according to our needs), and you may find its value here together with a solution in comments, we're done.

End of story.
Using W.A.as well as my previous calculations on harmonics sums.
 I find $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad -\frac{\pi^2G}{4}-2G+\frac{7}{4}\zeta(3)-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi}{2}-\frac{\pi^2}{4}+\frac{\pi^3}{16}+\frac{\pi^4}{64}+\ln2-2-3\beta(4)+\frac{\Psi^{(3)}(\frac{1}{4})}{256}- \frac{\Psi^{(3)}(\frac{3}{4})}
{256}$$
 $$ S=0,047743102114778065267...$$
 But I'm not sure of the result.