TITLE: How prove this sum $\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}$
QUESTION [26 upvotes]: 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

REPLY [5 votes]: $\newcommand{\+}{^{\dagger}}
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$\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}