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.

Solution 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

$$ E [2^{-\rho} ] ^2 -4E [ 2^{-\rho}]+1=0,$$

$$(E[ 2^{-\rho}] -2)^2-3=0 \quad \Rightarrow \quad E[2^{-\rho}] = 2-\sqrt{3} $$

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,

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

Solution 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.

Solution 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: