<s>[INST] Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$ 
Question: Show that 
  $$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$

From Wolfram alpha, it seems that the equality above is indeed correct.
But I do not know how to prove it.
Any hint is appreciated. 
 [/INST] Famously, $\displaystyle \int_0^{\pi/2} \cos^{2n}{x}\,\mathrm{d}x = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$ (e.g. see here); and $\displaystyle \frac{1}{1+n} = \int_0^1 y^n \, \mathrm{d} y$, thus:
$\displaystyle \begin{aligned} \sum_{n \ge 0} \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} & = \frac{1}{\pi} \sum_{n \ge 0}\int_0^{\pi/2}\int_0^1 y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} (y\cos^2{x})^n\,\mathrm{d}y\,\mathrm{d}x\,  \\& = \frac{1}{\pi}\int_0^{\pi/2}\int_0^1\frac{1}{1-y \cos^2{x}}\mathrm{d}y\,\mathrm{d}x \\& = \frac{1}{\pi} \int_0^{\pi/2}\sec^2{x} \cdot \log\left({\csc^2{x}}\right)\,\mathrm{d}x\, \\& = \frac{1}{\pi} \cdot \bigg[2x+\tan{x}\log(\sec^2{x})\bigg]_{x \to 0}^{x \to \pi/2} \\& = \frac{1}{\pi}\cdot \pi \\& = 1.  \end{aligned} $
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