<s>[INST] 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.
 [/INST] $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\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
$$
{1 \over \root{1 - 4x}} = \sum_{n = 0}^{\infty}x^{n}{2n \choose n}
\quad\imp\quad
 \color{#f00}{\sum_{n = 0}^{\infty}{1 \over 16^{n}}{2n \choose n}} =
\color{#f00}{{2 \over \root{3}}}
$$
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