<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] The Generalized Binomial Theorem says
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac12\frac{x}{1!}+\frac12\!\cdot\!\frac32\frac{x^2}{2!}+\frac12\!\cdot\!\frac32\!\cdot\!\frac52\frac{x^2}{3!}+\cdots\\[6pt]
&=\sum_{k=0}^\infty(2k-1)!!\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\infty\frac{(2k!)}{2^kk!}\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac x4\right)^k\tag1
\end{align}
$$
Substituting $x\mapsto x/4$ gives
$$
\frac1{\sqrt{1-4x}}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Integrating gives
$$
\frac12-\frac12\sqrt{1-4x}=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}x^{k+1}\tag3
$$
Plug in $x=\frac14$ and multiply by $2$
$$
1=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}\frac1{2^{2k+1}}\tag4
$$
</s>