Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$

Question

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.

Answer 1

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,

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

Answer 2

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.

Answer 3

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:

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

Answer 4

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

Answer 5

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}{1 \over 16^{n}}{2n \choose n}:\,?}$.

---

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 $$ {1 \over \root{1 - 4x}} = \sum_{n = 0}^{\infty}x^{n}{2n \choose n} \quad\imp\quad \color{#f00}{\sum_{n = 0}^{\infty}{1 \over 16^{n}}{2n \choose n}} = \color{#f00}{{2 \over \root{3}}} $$