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.

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 $$\sum_{n\geq0}\dbinom{2n}{n}\frac{1}{16^{n}}=\frac{2}{\sqrt{3}}.$$
