8
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Well, $4^n4^n=16^n$, and $\sum sin^{2n}(x)/4^n$ is a geometric series... $\endgroup$ – David C. Ullrich Jun 3 '16 at 14:50
  • $\begingroup$ In fact, Wolfram Alpha evaluates this series easily $\endgroup$ – Yuriy S Jun 3 '16 at 19:53
8
$\begingroup$

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,

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

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ @OlivierOloa: thank you, fixed. $\endgroup$ – Jack D'Aurizio Jun 3 '16 at 16:52
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.