TITLE: Harmonic Numbers series I
QUESTION [14 upvotes]: Can it be shown that
\begin{align}
\sum_{n=1}^{\infty} \binom{2n}{n} \ \frac{H_{n+1}}{n+1} \ \left(\frac{3}{16}\right)^{n} = \frac{5}{3} + \frac{8}{3} \ \ln 2 - \frac{8}{3} \ \ln 3
\end{align}
where $H_{n}$ is the Harmonic number and defined as
\begin{align}
H_{n} = \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt. 
\end{align}

REPLY [9 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\sum_{n = 1}^{\infty}{2n \choose n}\,
     {H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}
     ={5 \over 3} + {8 \over 3}\,\ln\pars{2} - {8 \over 3}\,\ln\pars{3}
     \approx {\tt 0.5854}:\ {\large ?}}$.

\begin{align}
\mbox{Note that}\quad
H_{n + 1}&=\int_{0}^{1}{1 - t^{n + 1} \over 1 - t}\,\dd t
=-\pars{n + 1}\int_{0}^{1}\ln\pars{1 - t}t^{n}\,\dd t
\end{align}

With $\ds{\mu \equiv {3 \over 16}}$:
\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}{2n \choose n}\,
{H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}}
=-1 -\int_{0}^{1}\ln\pars{1 - t}\color{#00f}{\sum_{n = 0}^{\infty}%
{2n \choose n}\,\pars{\mu t}^{n}}\,\dd t\tag{1}
\end{align}

Lets evaluate the "$\ds{\color{#00f}{\mbox{blue sum}}}$":
  \begin{align}&\color{#00f}{\sum_{n = 0}^{\infty}{2n \choose n}\,\pars{\mu t}^{n}}
=\sum_{n = 0}^{\infty}\pars{\mu t}^{n}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z}\sum_{n = 0}^{\infty}\bracks{%
\pars{1 + z}^{2}\mu t \over z}^{n}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{1 \over z}
{1 \over 1 - \pars{1 + z}^{2}\mu t/z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=-\,{1 \over \mu t}\oint_{\verts{z}\ =\ 1}
{1 \over z^{2} + \bracks{2 - 1/\pars{\mu t}}z + 1}\,{\dd z \over 2\pi\ic}
\end{align}
  The integrand has one pole inside the contour
  $\ds{\pars{~p \equiv {1 - 2\mu t - \root{1 - 4\mu t} \over 2\mu t}~}}$ such that:
  \begin{align}
&\color{#00f}{\sum_{n = 1}^{\infty}{2n \choose n}\,\pars{\mu t}^{n}}
=-\,{1 \over \mu t}\,{1 \over 2p + 2 - 1/\pars{\mu t}}
=-\,{1 \over 2p\mu t + 2\mu t - 1}
={1 \over \root{1 - 4\mu t}}
\end{align}

Replacing in $\pars{1}$:
\begin{align}&\!\!\!\!\!\!\!\!\!\!
\color{#c00000}{\sum_{n = 1}^{\infty}{2n \choose n}\,
{H_{n + 1} \over n + 1}\ \overbrace{\pars{3 \over 16}^{n}}
^{\ds{\color{#000}{\mu^{n}}}}}\ =\
-1 - \int_{0}^{1}{\ln\pars{1 - t} \over \root{1 - 4\mu t}}\,\dd t
=-1 - \int_{0}^{1}{\ln\pars{1 - t} \over \root{1 - 3t/4}}\,\dd t\tag{2}
\end{align}
The integral is easily evaluated with the change of variables
$\ds{x \equiv \root{1 - 3t/4}}$:
\begin{align}&\int_{0}^{1}{\ln\pars{1 - t} \over \root{1 - 3t/4}}\,\dd t
=-\,{8 \over 3}\int_{1}^{1/2}\bracks{\ln\pars{4 \over 3} + \ln\pars{x + \half}
+ \ln\pars{x - \half}}\,\dd x
\\[3mm]&={8 \over 3}\bracks{\ln\pars{3 \over 2} - 1}
\end{align}

By replacing this result in $\pars{2}$, we find:
  $$\color{#66f}{\large%
\sum_{n = 1}^{\infty}{2n \choose n}\,{H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}
={5 \over 3} + {8 \over 3}\,\ln\pars{2} - {8 \over 3}\,\ln\pars{3}}
\approx {\tt 0.5854}
$$