TITLE: Computing $\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$
QUESTION [10 upvotes]: An advanced sum proposed by Cornel Valean:

$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$$
$$=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^22\zeta(2)+6\ln2\zeta(2)+\frac16\ln^42-1$$

I managed to find the integral representation of $\ \displaystyle\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}\ $ but not $S$:
Since
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
we can write
$$\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{2^{2n}x^{n}}{n{2n\choose n}}$$
now multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we have
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\tag1$$
But I could not get the integral representation of $S$. Any idea?
In case you find the integral, I prefer solutions that do not use contour integration or you can leave it to me to give it a try. Thank you.
In case the reader is curious about computing the integral in $(1)$, set $x=\sin^2\theta$ then use the Fourier series of $\ln(\cos \theta)$.

REPLY [6 votes]: Following @Felix's idea above:
$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}=\sum_{n=2}^\infty\frac{2^{2n-2}H_n}{n^2{2n-2\choose n-1}}$$
Note that
$$\frac{{2n+2\choose n+1}}{{2n\choose n}}=\frac{\frac{\Gamma(2n+3)}{\Gamma^2(n+2)}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{\frac{(2n+2)(2n+1)\Gamma(2n+1)}{((n+1)\Gamma(n+1))^2}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{2(2n+1)}{n+1}$$
replace $n$ by $n-1$ we get
$$\frac{1}{{2n-2\choose n-1}}=\frac{2(2n-1)}{n{2n\choose n}}$$
Therefore
$$S=\sum_{n=2}^\infty\frac{2^{2n-1}(2n-1)H_n}{n^3{2n\choose n}}=\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}-\frac12 \sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^3 {2n\choose n}}-1\tag1$$
In the question body we have
$$\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2} \theta \ln(\cos\theta)d\theta$$
$$=-8\int_0^{\pi/2}\theta\left(-\ln(2)-\sum_{n=1}^\infty\frac{(-1)^n\cos(2n\theta)}{n}\right)d\theta=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag2$$
and here we already showed
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)\tag3$$
Finally, plug $(2)$ and $(3)$ in $(1)$ we obtain
$$S=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^2(2)\zeta(2)+6\ln(2)\zeta(2)+\frac16\ln^4(2)-1$$