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.

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, 
$$ S = \frac{2}{\sqrt{3}}.$$ 
