Studying on this sum interesting $\sum_{n=1}^{\infty}\frac{{2n \choose n }}{4^n n}$

Start off with the sum given in the comments and index the sum starting from $n=1$ to get that$$\sum\limits_{n\geq1}\binom {2n}n\left(\frac x4\right)^n=\frac 1{\sqrt{1-x}}-1$$Now divide both sides by $x$ and integrate from zero to one.$$\begin{align*}\sum\limits_{n\geq1}\binom {2n}n\frac 1{4^nn} & =\int\limits_0^1\mathrm dx\,\left[\frac 1{x\sqrt{1-x}}-\frac 1x\right]\\ & =\log 4\end{align*}$$

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Sequences And Series

Closed Form

Golden Ratio

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