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. 
 A: Probabilistically, the number
$$2\,C_n\,\left(\frac{1}{2}\right)^{2(n+1)}=\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}$$
is the probability that a symmetric random walk on the lattice points of $\mathbb{R}$ will return to the starting point for the first time after $2(n+1)$ steps.  However, it is not difficult to show that with probability $1$, this random walk returns to the starting point (see, for example, Theorem 3 of this link).  This shows that
$$\sum_{n=0}^\infty\,\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}=1.$$
The same idea can be used to verify the generating function of the Catalan numbers (by considering asymmetric random walks instead).
A: 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 $2$
$$
1=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}\frac1{2^{2k+1}}\tag4
$$
A: Let
$$ f(x)=\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} x^{2n+1} $$
and then
$$ (xf(x))'=2\sum_{n=0}^\infty \binom{2n}{n} x^{2n+1}=\frac{2x}{\sqrt{1-4x^2}}. $$
So
$$ xf(x)=\int_0^x\frac{2t}{\sqrt{1-4t^2}}dt=\frac12(1-\sqrt{1-4x^2})$$
and hence
$$ f(\frac12)=1$$
A: 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} $
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 0}^{\infty}{1 \over n + 1}{2n \choose n}
{1 \over 2^{2n + 1}} & =
\sum_{n = 0}^{\infty}{1 \over n + 1}\
\overbrace{2n \choose n}^{\ds{= {-1/2 \choose n}\pars{-4}^{n}}}
{1 \over 2^{2n + 1}}\label{1}\tag{1}
\end{align}
In (\ref{1}), I used a
known identity.
Then,
\begin{align}
\sum_{n = 0}^{\infty}{1 \over n + 1}{2n \choose n}
{1 \over 2^{2n + 1}} & =
{1 \over 2}\sum_{n = 0}^{\infty}{-1/2 \choose n}
{\pars{-1}^{n} \over n + 1}
\\[5mm] & =
{1 \over 2}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-1}^{n}
\int_{0}^{1}t^{n}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\sum_{n = 0}^{\infty}
{-1/2 \choose n}\pars{-t}^{n}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}{\dd t \over \root{1 - t}} =
\bbx{\large 1}
\end{align}
