<s>[INST] 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?
 [/INST] 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.
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