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?
 A: $$\color{royalblue}{\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n})}{n2^n \binom{2n}{n}} = \frac{\pi ^2}{36}-\frac{\log ^2(2)}{3}}$$
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:
$$\small\begin{aligned}\sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^22^n \binom{2n}{n}} &= -\frac{\zeta (3)}{8}-\frac{1}{6} \log ^3(2)+\frac{1}{12} \pi ^2 \log (2) \\ \sum \limits _{n=1}^{\infty}\frac{(-1)^{n-1} (H_{2n}-H_{n-1})}{n^32^n \binom{2n}{n}} &= -2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{13}{8} \zeta (3) \log (2)+\frac{19 \pi ^4}{720}-\frac{1}{24} \log ^4(2)+\frac{1}{24} \pi ^2 \log ^2(2) \end{aligned}$$
A: Simplify the integral
\begin{align}
&\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx\\
=& \>\sqrt2 \left(\sinh^{-1}\frac x{\sqrt8}\right)^2\bigg|_0^1 
 - \int_0^1\frac{\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,\overset{x=\sqrt2(y-\frac1y)}{dx}\\
=&\>\frac1{2\sqrt2}\ln^22-2\sqrt2 \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\tag1
\end{align}
With $\int_0^1 \frac{\ln t}{1+t}dt=-\frac{\pi^2}{12}$ and $\int_0^1 \frac{\ln t}{1-t}dt=-\frac{\pi^2}{6}$
\begin{align}\int_0^{1/2} \frac{\ln t}{1-t}dt\overset{ibp}=&
-\ln^22+\int_0^{1/2} \frac{\ln (1-t)}{t}\overset{t\to1-t}{dt}\\
=&\>\frac12\left(-\ln^22+\int_0^{1} \frac{\ln t}{1-t} dt\right)
=-\frac12\ln^22 -\frac{\pi^2}{12}
\end{align}
\begin{align}
& \int_1^{\sqrt2}\frac {\ln y}{\sqrt2 y^2+y-\sqrt2 }dy\\
=& \>\frac13\bigg(\int_1^{\sqrt2}\frac{\ln y}{y-\frac1{\sqrt2}}\overset{y=\frac1{\sqrt2t}}{dy}- \int_1^{\sqrt2} \frac{\ln y}{y+{\sqrt2}} \overset{y={\sqrt2t}}{dy}\bigg)\\
=&\>\frac13\bigg( -\frac38\ln^22-\int_0^1 \frac{\ln t}{1+t}dt+\int_0^{1/2}\frac{\ln t}{1-t}dt-\int_0^{1/\sqrt2} \frac{2t\ln t}{1-t^2}\overset{t^2\to t}{dt} \bigg)\\
=&\>-\frac18\ln^22+\frac{\pi^2}{36}
+\frac16 \int_0^{1/2}\frac{\ln t}{1-t}dt
= \frac{\pi^2}{72}-\frac5{24}\ln^22
\end{align}
Plug into (1) to obtain
$$\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx=\frac{2\sqrt2}{3}\left(\ln^22-\frac{\pi^2}{24}\right)$$
A: After some rather lengthy yet not particularly difficult manipulations I have concluded that the more general series can be expressed as
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}(H_{2n}-H_{n})}{n2^n\binom{2n}{n}}x^{2n}=\frac{8x}{\sqrt{8+x^2}}\int_0^x\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt,~~~\lvert x\rvert\leqslant2\sqrt2.$$
Hence, we may evaluate your given series by plugging in $x=1$, producing the result
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}(H_{2n}-H_{n})}{n2^n\binom{2n}{n}}=\frac{8}{3}\int_0^1\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt.$$
This integral may be easier to work with than the integral that you arrived at, but at this moment in time I am unsure how to manipulate it myself. I have managed to show that
$$\begin{align}\frac{8}{3}\int_0^1\frac{\operatorname{arsinh}^2\frac{t}{2\sqrt2}}{t^2\sqrt{8+t^2}}dt&=-\frac{1}{4}\log^22+\frac{2}{3}\int_0^1\frac{\operatorname{arsinh}\frac{t}{2\sqrt2}}{t}dt\\
&=-\frac{1}{4}\log^22-\frac{4\sqrt2}{3}\int_0^1\frac{\log t}{\sqrt{8+t^2}}dt\end{align}$$ using integration by parts but I'm not really sure where to go from there.

If you would like to know how to obtain these integrals then please post a comment. I'm afraid it's quite long so I won't show my full working in all likelihood.

I hope that you find this helpful. If you have any questions then please don't hesitate to ask.
