How prove this sum $\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}$ show that
$$\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}=5+4\sqrt{2}\left(\log{\dfrac{2\sqrt{2}}{1+\sqrt{2}}}-1\right)$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My try: we let
$$s(x)=\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{(n+1)}x^{n+1}$$
then
$$s'(x)=\sum_{n=1}^{\infty}(-1)^{n-1}\binom{2n}{n}H_{n+1}x^n$$
then I can't.Thank you
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}{2n \choose n}{\pars{-1}^{n - 1}H_{n + 1}
     \over 4^{n}\pars{n + 1}}
     = 5 + 4\root{2}\bracks{\ln\pars{2\sqrt{2} \over 1 + \root{2}} - 1}
     \approx {\tt 0.2389}:\ {\large ?}}$

\begin{align}
\mbox{Note that}\quad\sum_{n = 1}^{\infty}
{2n \choose n}{\pars{-1}^{n - 1}H_{n + 1} \over 4^{n}\pars{n + 1}}=
-\sum_{n = 1}^{\infty}{2n \choose n}{H_{n + 1} \over n + 1}\,
\pars{-\,{1 \over 4}}^{n}\tag{1}
\end{align}

In a previous answer, I did a detailed derivation of:
$$
\sum_{n = 1}^{\infty}{2n \choose n}{H_{n + 1} \over n + 1}\,\mu^{n}
=-1 - \int_{0}^{1}{\ln\pars{1 - t} \over \root{1 - 4\mu t}}\,\dd t
$$

such that $\pars{1}$ is reduced to:
  \begin{align}&\color{#66f}{\large\sum_{n = 1}^{\infty}
{2n \choose n}{\pars{-1}^{n - 1}H_{n + 1} \over 4^{n}\pars{n + 1}}}=
1 +\
\overbrace{\int_{0}^{1}{\ln\pars{1 - t} \over \root{1 + t}}\,\dd t}
^{\ds{\mbox{Set}\  x \equiv \root{1 + t}}}
\\[3mm]&=1 + 2\int_{1}^{\root{2}}
\bracks{\ln\pars{\root{2} + x} + \ln\pars{\root{2} - x}}\,\dd x
\\[3mm]&=\color{#66f}{\large%
5 + 4\root{2}\bracks{\ln\pars{2\sqrt{2} \over 1 + \root{2}} - 1}}
\approx {\tt 0.2389}
\end{align}

A: There's a transformation formula (might have been derived from generalized Euler series transformation):
$$
f(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, a_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, g\left(\dfrac{x}{1+4\, x}\right)
$$
where
$$a_n=\sum_{k=0}^n\, \binom{n}{k}\, (-1)^{n-k}\, b_k$$
and
$$g(x)=\sum_{n=0}^\infty \, \binom{2n}{n}\, b_n \, x^n$$
For $a_n=(-1)^{n-1}\, H_n$ and $b_n=\dfrac{1}{n}$, we have the identity
$$
(-1)^{n-1}\, H_n = \sum_{k=1}^n \binom{n}{k}\, \dfrac{(-1)^{n-k}}{n}
$$
Also, from the generating function for central binomial coefficients, the following g.f. can be obtained:
$$
G_1(x)=\sum_{n=1}^\infty \, \binom{2n}{n}\, \dfrac{1}{n}\, x^n = 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, x}}\right)}
$$
Therefore,
$\displaystyle \begin{aligned}
f(x)&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, H_n \, x^n = \dfrac{1}{\sqrt{1+4\, x}}\, \sum_{n=1}^\infty \, 
\binom{2n}{n} \, \dfrac{1}{n}\, \left( \dfrac{x}{1+4\, x} \right)^n\\\\
&= \dfrac{1}{\sqrt{1+4\, x}}\, G_1\left(\dfrac{x}{1+4\, x}\right)\\\\
&= \dfrac{1}{\sqrt{1+4\, x}}\, 2\, \log{\left( \dfrac{2}{1+\sqrt{1-4\, \left(\dfrac{x}{1+4\, x}\right)}}\right)}
\end{aligned}
$
Then, consider the g.f. we need in order to get the required sum:
$\displaystyle \begin{aligned}
h(x) &=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n+1}}{n+1} \, x^n\\\\
&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}+\dfrac{1}{n+1}}{n+1} \, x^n\\\\
&=\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{H_{n}}{n+1} \, x^n+\sum_{n=1}^\infty \, \binom{2n}{n}\, (-1)^{n-1}\, \dfrac{1}{(n+1)^2} \, x^n\\\\
&=h_1(x)+h_2(x)
\end{aligned}
$
Both $h_1(x)$ and $h_2(x)$ can be found by integrating the g.f.'s we have already seen:
$\displaystyle \begin{aligned}
h_1(x)&=\dfrac{1}{x}\int f(x) \, dx+\dfrac{4 \, \log\left(2\right) + 1}{4 \, x}\\\\
&=\dfrac{4 \, \sqrt{4 \, x + 1} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{4 \, x + 1}} + 1}\right) - 4 \, \log\left(\sqrt{4 \, x + 1} + 1\right) - 1}{4 \, x} + \dfrac{4 \, \log\left(2\right) + 1}{4 \, x}
\end{aligned}$
$\displaystyle
\begin{aligned}
h_2(x)&= \dfrac{1}{x}\, \int \left(\dfrac{1}{x}\, \int -\dfrac{1}{\sqrt{1+4\, x}}\, dx \right)\, dx + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1\\\\
&= -\dfrac{2 \, \sqrt{4 \, x + 1} - \log\left(\sqrt{4 \, x + 1} + 1\right) + \log\left(\sqrt{4 \, x + 1} - 1\right)}{2 \, x} + \dfrac{\log\left(x\right) + 2}{2 \, x} + 1
\end{aligned}
$
and the required sum is:
$\displaystyle \begin{aligned} h\left(\dfrac{1}{4}\right)&=4 \, \sqrt{2} \log\left(\dfrac{2}{\sqrt{\dfrac{1}{2}} + 1}\right) - 4 \, \sqrt{2} + 4 \, \log\left(2\right) + 2 \, \log\left(\dfrac{1}{4}\right) - 2 \, \log\left(\sqrt{2} + 1\right) - 2 \, \log\left(\sqrt{2} - 1\right) + 5\\\\
&=4 \, \sqrt{2} {\left(\log\left(\dfrac{2 \, \sqrt{2}}{\sqrt{2} + 1}\right) - 1\right)} + 5\\\\
&\approx 0.238892690197059
\end{aligned}$
References: 
[1]
[2]
