Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$ 
Prove that
  $$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$

According to Wolfram the above holds. Could someone show me the steps for this?
 A: HINT
First note that
$$\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac{(n+1)-1}{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}-\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}$$
Now recall that
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n}=\frac1{\sqrt{1-x}},~~~\text{for }|x|<1$$
The task boils down to an exercise in integration. Can you take it from here?

For the interested reader I will add the complete solution now. As given, note that for the first sum we have the following
$$\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}=\sum_{n\geq0}\binom{2n}n\frac1{4^n}\int_0^1x^n{\rm d}x=\int_0^1\frac{{\rm d}x}{\sqrt{1-x}}=[-2\sqrt{1-x}]_0^1=2$$
This also gives us the following identity
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n(n+1)}=\int\frac{{\rm d}x}{\sqrt{1-x}}=-2\sqrt{1-x}+c$$
Take $x=0$ to see that $c=2$. Dividing by $x$ and integrating again over $[0;1]$ gives
\begin{align*}
\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}&=\int_0^1\frac{2-2\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt x}{1-x}{\rm d}x\\
&=2\int_0^1\frac1{1+\sqrt x}{\rm d}x\\
&=2\sum_{n\geq0}(-1)^n\int_0^1x^{\frac n2}{\rm d}x\\
&=2\sum_{n\geq0}\frac{(-1)^n}{\frac n2+1}\\
&=4\sum_{n\geq0}\frac{(-1)^n}{n+2}\\
&=4\left[\sum_{n\geq1}\frac{(-1)^n}n+1\right]\\
&=4-4\log2
\end{align*}
The result follows.

$$\therefore~\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}~=~4\log2-2$$

A: The $n$-th Catalan number is defined as
$$ C_n := \frac 1 {n+1}\binom{2n}n.$$
By inspecting the family of definite integrals
$$ J_n(\alpha) := \frac 1 \pi \int_0^\alpha \xi^{2n} \sqrt{\alpha^2-\xi^2} d\xi, \qquad \alpha > 0,\ n\geq 0, $$
it can be shown, via standard integration techniques, that $C_n = J_n(2)$:
equivalently, through the substitution $x=\xi^2$, $$ C_n = \frac{1}{2\pi} \int_0^4 x^n \sqrt{\frac{4-x}{x}} dx.$$
Then your sum $S$ becomes
$$\begin{split}
S &=\sum_{n=0}^\infty \frac{ n C_n}{(n+1) 4^n } = \sum_{n=0}^\infty \frac{(n+1-1) C_n}{(n+1)4^n} = \sum_{n=0}^\infty \frac{C_n}{4^n} - \sum_{n=0}^\infty \frac{C_n}{4^n(n+1)} \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n -\sum_{n=0}^\infty \frac{(x/4)^n}{n+1} \right] dx \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{1}{1-x/4} + \frac{\log(1-x/4)}{x/4} \right]dx \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{4}{4-x}+\frac 4 x \log\left(\frac{4-x}{4} \right) \right] dx \\
&=\frac{2}{\pi} \int_0^4 \frac{dx}{\sqrt{4-(x-2)^2}} + \frac 2 \pi \int_0^4 \frac 1 x \log\left(\frac{4-x}{4} \right) dx \\
&= \frac 2\pi \int_0^\pi  dt + \frac{2}{\pi} \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds \\
&=: 2+\frac 2\pi I,
\end{split}$$
where we have interchanged sum and integral by total convergence of power series, and employed the formula for geometric series and its integral, and parallel substitutions $x = 2-2\cos t$, $s = 1 - x/4$. The definite integral $I$ can be resolved as follows:
$$\begin{split}
I &= \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds=\int_0^\infty \log \left( \frac y {y+1}\right) \frac{\sqrt y}{y+1} dy = 2\int_0^\infty \log \left( \frac {x^2} {x^2+1}\right) \frac{x^2}{x^2+1} dx \\
&= \underbrace{\left[(x-\arctan x)\log \left( \frac {x^2} {x^2+1}\right)  \right]_{-\infty}^{+\infty}}_{=0} -2 \int_{-\infty}^\infty \frac{x-\arctan x}{x(1+x^2)}dx \\
&= -2\pi +2 \int_{-\pi/2}^{\pi/2} z \cot(z) dz = -2\pi + 4 \left(\underbrace{[z \log(\sin z)]_0^{\pi/2}}_{=0} - \int_0^{\pi/2} \log(\sin z) dz \right),
\end{split}$$
where we have employed substitutions $y = s/(1-s)$, $x = \sqrt y$, and then $z = \arctan x$. The remaining integral has been evaluated many times over on MSE (see this answer): in total, we have $$I = 2\pi(\log 2 - 1), \quad \implies \quad S = 2 + 4 (\log 2 -1) = \boxed{\log(16)-2} $$
as required.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{%
\sum_{n = 0}^{\infty}{2n \choose n}{n \over 4^{n}\pars{n + 1}^{2}}} =
\sum_{n = 0}^{\infty}\bracks{{-1/2 \choose n}\pars{-4}^{n}}{1 \over 4^{n}}
\bracks{{1 \over n + 1} - {1 \over \pars{n + 1}^{2}}}
\\[5mm] = &
\left.\pars{1 + \partiald{}{a}}\sum_{n = 0}^{\infty}{-1/2 \choose n}
{\pars{-1}^{n} \over n + a + 1}\,\right\vert_{\large\ a\ =\ 0}
\\[5mm] = &
\left.\pars{1 + \partiald{}{a}}\sum_{n = 0}^{\infty}{-1/2 \choose n}
\pars{-1}^{n}\int_{0}^{1}t^{n + a}\,\dd t\,\right\vert_{\large\ a\ =\ 0}
\\[5mm] = &
\left.\pars{1 + \partiald{}{a}}\int_{0}^{1}t^{a}
\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-t}^{n}\,\dd t\,\right\vert_{\large\ a\ =\ 0}
\\[5mm] = &\
\left.\pars{1 + \partiald{}{a}}\int_{0}^{1}t^{a}\pars{1 - t}^{-1/2}
\,\dd t\,\right\vert_{\large\ a\ =\ 0}
\\[5mm] = &\
\left.\pars{1 + \partiald{}{a}}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}\,\right\vert_{\large\ a\ =\ 0}
\\[5mm] = &\
\underbrace{{\Gamma\pars{1}\Gamma\pars{1/2} \over \Gamma\pars{3/2}}}
_{\ds{2}}\ +\
\underbrace{\left.\partiald{}{a}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}\,\right\vert_{\large\ a\ =\ 0}}
_{\ds{-4 + \ln\pars{16}}}
\\[5mm] = &\
\bbx{\ln\pars{16} - 2} \approx 0.7726 \\ &
\end{align}

$\underline{Note\ that}$
\begin{align}
&\bbox[5px,#ffd]{\left.\partiald{}{a}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}
\,\right\vert_{\ a\ =\ 0}} =
\bracks{a^{1}}{\Gamma\pars{1 + a}\Gamma\pars{1/2} \over \Gamma\pars{3/2 + a}}
\\[5mm] = &\
\Gamma\pars{1 \over 2}\bracks{a^{1}}
{\Gamma\pars{1} + \Gamma'\pars{1}a \over
\Gamma\pars{3/2} +  \Gamma'\pars{3/2}a} =
\Gamma\pars{1 \over 2}{1 \over \Gamma\pars{3/2}}\bracks{a^{1}}
{1 + \Psi\pars{1}a \over 1 +  \Psi\pars{3/2}a}
\\[5mm] = &\
2\bracks{a^{1}}
\pars{1 - \gamma a}\bracks{1 -  \Psi\pars{3 \over 2}a} =
-2\bracks{\gamma + \Psi\pars{1 \over 2} + 2}
\\[5mm] = &\
-4 - 2\int_{0}^{1}{1 - t^{-1/2} \over 1 - t}\,\dd t =
-4 - 2\int_{0}^{1}{1 - t^{-1} \over 1 - t^{2}}\,2t\,\dd t =
-4 + 4\int_{0}^{1}{\dd t \over 1 + t}
\\[5mm] = &\
-4 + 4\ln\pars{2} = \bbx{-4 + \ln\pars{16}} \\ &
\end{align}
