Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$ 
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$

$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
 A: Here's another probabilistic approach. We don't evaluate any (Riemann) integrals. 
Consider a simple symmetric random walk on ${\mathbb Z}$ starting from $0$ at time $0$. The probability that at time $2n$ the walk is at zero is equal to $\binom{2n}{n}2^{-2n}$, and the probability that at time $2n+1$ the walk is at zero is $0$. 
From this it follows  that the expression we wish to evaluate is 
$$S=  \sum_{j=0}^\infty E[  2^{-T_j}].$$ 
where $0=T_0<T_1<\dots $ are the times the walk is at zero. Note that $(T_{j+1}-T_j)$ are IID and have the same distribution as $T_1$. Therefore, this is a geometric series. Its sum is 
$$S = \frac{1}{1-E [2^{-T_1}]}.$$ 
To compute $E [ 2^{-T_1}]$ we consider first $\rho$, the time until the walk hits $1$, starting from $0$. Conditioning on the first step, we have 
$$ E[ 2^{-\rho}] = 2^{-1} \left ( \frac 12 + \frac 12 E [2^{-\rho}]^2\right),$$ 
representing, either moving to the right first or moving to the left first. Therefore 
$$ E [2^{-\rho} ] ^2 -4E [ 2^{-\rho}]+1=0,$$ 
or 
$$(E[ 2^{-\rho}] -2)^2-3=0 \quad \Rightarrow \quad E[2^{-\rho}] = 2-\sqrt{3} $$
Let's go back to our original problem. Conditioning  on the first step,
$$ E [2^{-T_1} ] = 2^{-1} E [ 2^{-\rho}]=1- \frac{\sqrt{3}}{2}.$$ 
Thus, 
$$ S = \frac{2}{\sqrt{3}}.$$ 
A: Hint. From what you have proved, one may deduce that
$$
\frac{\pi}{2}\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}=\int_{0}^{\large \frac{\pi}{2}}\sum^{\infty}_{n=0}\frac{1}{4^n}\sin^{2n}xdx=\int_{0}^{\large \frac{\pi}{2}}\frac{4}{4-\sin^2 x}dx
$$ the latter integral being easy to evaluate.
A: Just completing Olivier's answer:
$$ I=\int_{0}^{\pi/2}\frac{4\,dx}{4-\sin^2 x}=\int_{0}^{\pi/2}\frac{4\,dx}{4-\cos^2 x}=\int_{0}^{+\infty}\frac{dt}{1+t^2}\cdot\frac{4}{4-\frac{1}{1+t^2}}$$
gives:

$$ S = \sum_{n\geq 0}\frac{1}{16^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{4}{4t^2+3} = \color{red}{\frac{2}{\sqrt{3}}}.$$

A: Maybe it is interesting to see that $$S=\sum_{n\geq0}\dbinom{2n}{n}x^{n}=\sum_{n\geq0}\frac{\left(2n\right)!}{n!^{2}}x^{n}$$ and now note that $$\frac{\left(2n\right)!}{n!}=2^{n}\left(2n-1\right)!!=4^{n}\left(\frac{1}{2}\right)_{n}=\left(-1\right)^{n}4^{n}\left(-\frac{1}{2}\right)_{n}
 $$ where $(x)_{n}$ is the Pochhamer' symbol. So we have, by the generalized binomial theorem, that $$S=\sum_{n\geq0}\dbinom{-1/2}{n}\left(-1\right)^{n}4^{n}x^{n}=\frac{1}{\sqrt{1-4x}},\left|x\right|<\frac{1}{4}$$ so if we take $x=\frac{1}{16}$ we have $$\sum_{n\geq0}\dbinom{2n}{n}\frac{1}{16^{n}}=\frac{2}{\sqrt{3}}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\sum_{n = 1}^{\infty}{1 \over 16^{n}}{2n \choose n}:\,?}$.


Lets $\ds{\mathrm{f}\pars{x} \equiv \sum_{n=0}^{\infty}{2n \choose n}x^{n}}$. Then,
\begin{align}
\mathrm{f}\pars{x} & =
1 + \sum_{n = 1}^{\infty}x^{n}\,{2n\pars{2n - 1}\pars{2n - 2}! \over
n\pars{n - 1}!\,n\pars{n - 1}!} =
1 + 2\sum_{n = 1}^{\infty}x^{n}\,{2n - 1 \over n}{2n - 2 \choose n - 1}
\\[3mm] & =
1 + 2\sum_{n = 0}^{\infty}x^{n + 1}\,{2n + 1 \over n + 1}{2n \choose n} =
1 + 4x\,\mathrm{f}\pars{x} - 2\sum_{n = 0}^{\infty}x^{n + 1}\,
{1 \over n + 1}{2n \choose n}
\\[3mm] & =
1 + 4x\,\mathrm{f}\pars{x} - 2\sum_{n = 0}^{\infty}x^{n + 1}\,
{2n \choose n}\int_{0}^{1}y^{\,n}\,\dd y =
1 + 4x\,\mathrm{f}\pars{x} - 2x\int_{0}^{1}\,\mathrm{f}\pars{xy}\,\dd y
\\[3mm] & =
1 + 4x\,\mathrm{f}\pars{x} - 2\int_{0}^{x}\,\mathrm{f}\pars{y}\,\dd y
\\[8mm] 
\imp\quad\,\mathrm{f}'\pars{x} & = 4\,\mathrm{f}\pars{x} + 4x\,\mathrm{f}'\pars{x} - 2\,\mathrm{f}\pars{x}
\end{align}

$\ds{\mathrm{f}\pars{x}}$ satisfies:
$$
\mathrm{f}'\pars{x} - {2 \over 1 - 4x}\,\,\mathrm{f}\pars{x} = 0\,,
\qquad\,\mathrm{f}\pars{0} = 1
$$
Moreover,
$$
0 =\totald{\bracks{\root{1 - 4x}\,\mathrm{f}\pars{x}}}{x} =0\quad\imp
\root{1 - 4x}\,\mathrm{f}\pars{x} = \root{1 - 4 \times 0}\,\mathrm{f}\pars{0}
= 1
$$
such that
$$
{1 \over \root{1 - 4x}} = \sum_{n = 0}^{\infty}x^{n}{2n \choose n}
\quad\imp\quad
 \color{#f00}{\sum_{n = 0}^{\infty}{1 \over 16^{n}}{2n \choose n}} =
\color{#f00}{{2 \over \root{3}}}
$$
