I would like to know if it is possible to find a closed form of the sum $$\sum_{n=1}^{\infty} \frac{{H_{n-1}^{(2)}}x^{2n}}{n^2{{2n}\choose{n}}}.$$
Here $H_{n}^{(2)}$ denotes the generalized harmonic number.
This series look similar to the $\arcsin(x)$ series
$$\sum_{n=1}^{\infty} \frac{x^{2n}}{n^2{{2n}\choose{n}}}$$ but I don't know how to combine it with harmonic number. I tried converting it to Beta function but that didn't seem to help much.
I am interested in an even more general question, i.e. finding the closed form of $$\sum_{n=1}^{\infty} \frac{{H_{n-1}^{(k)}}x^{2n}}{n^2{{2n}\choose{n}}}$$ for any positive integer $k$. Any help is very appreciated.
The result $$\sum _{n=1}^{\infty } \frac{\sum _{k=1}^{n-1} \frac{1}{k^2} }{n^2 \binom{2 n}{n}}\,x^{2 n}=\frac{2}{3} \arcsin^4 \left(\frac{x}{2}\right)$$ is obtained by entering a slightly modified latex code into Wolfram Alpha, for example like this (I don't know why this variant of Latex code works and other equivalent ones don't).
The only way I know of demonstrating this, is by already knowing the result and tediously evaluating the Cauchy Product for $\frac{1}{6}\left[2 \arcsin^2\left(\frac{x}{2}\right)\right]^2$, that is
$$\frac{2}{3} \arcsin^4\left(\frac{x}{2}\right)=\frac{1}{6} \left(\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^2 \binom{2 n}{n}}\right)^2=\frac{1}{6} \sum _{k=2}^{\infty } \left(\sum _{j=1}^{k-1} \frac{x^{2 k}}{\left(j^2 \binom{2 j}{j}\right) \left((k-j)^2 \binom{2 (k-j)}{k-j}\right)}\right)$$
Perhaps someone else can devise a neater less tedious proof of this result.