Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to expand the numerator (Harmonic numbers) in terms of integral, to get:
$\displaystyle \tag*{} \sum \limits_{n=1}^{\infty} \frac{(-1)^n}{n2^n\binom{2n}{n}} \int _{0}^{1} \frac{x^n - x^{2n}}{1-x} \ \mathrm dx$
And I found that with the help of series expansion of $\sin^{-1}(x)$ and subsituting $x = i \sqrt{x} /8 $ where $i^2=1$
$\displaystyle \tag*{} -2(\sinh^{-1} (\sqrt{x}/8))^2 = \sum \limits_{n=1}^{\infty} \frac{(-1)^nx^n}{n^22^n \binom{2n}{n}} $
But this has $n^2$ in the denominator, which makes it complicated. EDIT: we can eliminate $n^2$ by differentiating and multiplying by $x$ as mentioned in the comments. But now, how can we solve our sum since $H_{2n}-H_n$ is numerator?
And I have the general formula for generating sum:
$ \displaystyle \tag*{} \sum \limits _{n=1}^{\infty} \frac{x^n}{n^y \binom{2n}{n}}$
And this doesn't have $n$ in the denominator and also it has closed-form $\forall \ y \geq 2$
Maybe if there is a way of expressing the denominator in the form integral, the sum can be changed in evaluating the double integral. I think there are other easy ways (such as using Hypergeometric functions)? Any help would be appreciated.
EDIT 2:
From the help of comments and a quora user,
$\DeclareMathOperator{\arcsinh}{arcsinh}$
By @Bertrand87, we have:
$\displaystyle \tag{1} H_{k} - H_{2k} + \ln (2) = \int _{0}^{1} \frac{x^{2k}}{1+x} \ \mathrm dx$
To make use of this, we express our sum as follows:
$\displaystyle \tag*{} S = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}}  + \sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}}$
We know
$\displaystyle \tag*{} 2\arcsin^2(x) = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k}}{k^2 \binom{2k}{k}}$
We differentiate both sides w.r.t $x$ both sides,
$\displaystyle \tag*{} \frac{2 \arcsin(x)}{\sqrt{1-x^2}} = \sum \limits_{k=1}^{\infty} \frac{(2x)^{2k-1}}{k \binom{2k}{k}}$
Now, we multiply both sides by $(2x)$ and define $x:= ix/ \sqrt{8}$ to get:
$\displaystyle \tag{2} \frac{-2x \arcsinh (x/ \sqrt {8})}{\sqrt{8}\sqrt{1+x^2/8}} = \sum \limits_{k=1}^{\infty} \frac{(-1)^k x^{2k}}{k2^k \binom{2k}{k}}$
We now multiply both sides by $-1/(1+x)$ and integrate from $0$ to $1$ and arrive at:
$\displaystyle \tag*{} \frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx = \sum \limits_{k=1}^{\infty} \frac{(-1)^k(H_{2k}-H_k - \ln2)}{k2^k \binom{2k}{k}}$
Similarly, from $(2)$ if we let $x=1$ and multiply both sides by $\ln 2$, it yields:
$\displaystyle \tag*{} \frac{-2 \arcsinh (1/ \sqrt{8}) \ln 2}{ \sqrt{8} \sqrt{1 + 1/8}} =\sum \limits_{k=1}^{\infty} \frac{(-1)^k(\ln2)}{k2^k \binom{2k}{k}} \approx -0.1601$
Now, our only problem is to evaluate the integral:
$\displaystyle \tag*{} \boxed{\frac{2}{\sqrt {8}}\int_{0}^{1}\frac{x\arcsinh(x/ \sqrt{8})}{\sqrt{1+x^2/8} (1+x)} \ \mathrm dx} $
Can anyone help me with this integral?

$$\color{royalblue}{\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n})}{n2^n \binom{2n}{n}} = \frac{\pi ^2}{36}-\frac{\log ^2(2)}{3}}$$
Proof sketch: Suffices to find $S=\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n2^n \binom{2n}{n}}$. Note that
$$\int_0^1 \frac{x^n(1-x)^n}{x} \log x dx = \frac{1}{n\binom{2n}{n}} (H_{n-1}-H_{2n})$$
so $$S = \int_0^1 \frac{(1-x)\log x}{1+\frac{1}{2}x(1-x)} dx$$
whose antiderivative can be expressed via dilogarithm $\text{Li}_2$, using special value of $\text{Li}_2(1/2)$ completes the proof.

As direct generalizations:
$$\small\begin{aligned}\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^22^n \binom{2n}{n}} &= -\frac{\zeta (3)}{8}-\frac{1}{6} \log ^3(2)+\frac{1}{12} \pi ^2 \log (2) \\ \sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^32^n \binom{2n}{n}} &= -2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13}{8} \zeta (3) \log (2)+\frac{19 \pi ^4}{720}-\frac{1}{24} \log ^4(2)+\frac{1}{24} \pi ^2 \log ^2(2) \end{aligned}$$
