Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$

Question

Question: Show that $$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$

From Wolfram alpha, it seems that the equality above is indeed correct.

But I do not know how to prove it. Any hint is appreciated.

Answer 1

Famously, $\displaystyle \int_0^{\pi/2} \cos^{2n}{x}\,\mathrm{d}x = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$ (e.g. see here); and $\displaystyle \frac{1}{1+n} = \int_0^1 y^n \, \mathrm{d} y$, thus:

$\displaystyle \begin{aligned} \sum_{n \ge 0} \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} & = \frac{1}{\pi} \sum_{n \ge 0}\int_0^{\pi/2}\int_0^1 y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} (y\cos^2{x})^n\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi}\int_0^{\pi/2}\int_0^1\frac{1}{1-y \cos^2{x}}\mathrm{d}y\,\mathrm{d}x \\& = \frac{1}{\pi} \int_0^{\pi/2}\sec^2{x} \cdot \log\left({\csc^2{x}}\right)\,\mathrm{d}x\, \\& = \frac{1}{\pi} \cdot \bigg[2x+\tan{x}\log(\sec^2{x})\bigg]_{x \to 0}^{x \to \pi/2} \\& = \frac{1}{\pi}\cdot \pi \\& = 1. \end{aligned} $

Answer 2

Probabilistically, the number $$2\,C_n\,\left(\frac{1}{2}\right)^{2(n+1)}=\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}$$ is the probability that a symmetric random walk on the lattice points of $\mathbb{R}$ will return to the starting point for the first time after $2(n+1)$ steps. However, it is not difficult to show that with probability $1$, this random walk returns to the starting point (see, for example, Theorem 3 of this link). This shows that $$\sum_{n=0}^\infty\,\frac{1}{n+1}\,\binom{2n}{n}\,\frac{1}{2^{2n+1}}=1.$$ The same idea can be used to verify the generating function of the Catalan numbers (by considering asymmetric random walks instead).

Answer 3

The Generalized Binomial Theorem says $$ \begin{align} (1-x)^{-1/2} &=1+\frac12\frac{x}{1!}+\frac12\!\cdot\!\frac32\frac{x^2}{2!}+\frac12\!\cdot\!\frac32\!\cdot\!\frac52\frac{x^2}{3!}+\cdots\\[6pt] &=\sum_{k=0}^\infty(2k-1)!!\frac{x^k}{2^kk!}\\ &=\sum_{k=0}^\infty\frac{(2k!)}{2^kk!}\frac{x^k}{2^kk!}\\ &=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac x4\right)^k\tag1 \end{align} $$ Substituting $x\mapsto x/4$ gives $$ \frac1{\sqrt{1-4x}}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2 $$ Integrating gives $$ \frac12-\frac12\sqrt{1-4x}=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}x^{k+1}\tag3 $$ Plug in $x=\frac14$ and multiply by $2$ $$ 1=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}\frac1{2^{2k+1}}\tag4 $$

Answer 4

Let $$ f(x)=\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} x^{2n+1} $$ and then $$ (xf(x))'=2\sum_{n=0}^\infty \binom{2n}{n} x^{2n+1}=\frac{2x}{\sqrt{1-4x^2}}. $$ So $$ xf(x)=\int_0^x\frac{2t}{\sqrt{1-4t^2}}dt=\frac12(1-\sqrt{1-4x^2})$$ and hence $$ f(\frac12)=1$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 0}^{\infty}{1 \over n + 1}{2n \choose n} {1 \over 2^{2n + 1}} & = \sum_{n = 0}^{\infty}{1 \over n + 1}\ \overbrace{2n \choose n}^{\ds{= {-1/2 \choose n}\pars{-4}^{n}}} {1 \over 2^{2n + 1}}\label{1}\tag{1} \end{align} In (\ref{1}), I used a known identity.

Then, \begin{align} \sum_{n = 0}^{\infty}{1 \over n + 1}{2n \choose n} {1 \over 2^{2n + 1}} & = {1 \over 2}\sum_{n = 0}^{\infty}{-1/2 \choose n} {\pars{-1}^{n} \over n + 1} \\[5mm] & = {1 \over 2}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-1}^{n} \int_{0}^{1}t^{n}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\sum_{n = 0}^{\infty} {-1/2 \choose n}\pars{-t}^{n}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}{\dd t \over \root{1 - t}} = \bbx{\large 1} \end{align}