How to compute a $\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}$

Question

I need help with the following excersise:

Evaluate $$\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}\\\text{Hint: Use identity}\int_0^{\pi/2}{\sin^{2k+1}x\;dx}=\frac{2^{2k}k!^2}{(2k+1)!}$$

My attempt:

$$\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}=\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){\frac{2n!}{(2n-n)!n!}}}}\\=\sum_{n=0}^{\infty}{\frac{2^nn!^2}{(2n+1)!}}=\sum_{n=0}^{\infty}{\frac{2^n2^nn!^2}{2^n(2n+1)!}}=\sum_{n=0}^{\infty}{\frac{2^{2n}n!^2}{2^n(2n+1)!}}$$

Applying the identity

$$\sum_{n=0}^{\infty}{\frac{2^{2n}n!^2}{2^n(2n+1)!}}=\sum_{n=0}^\infty{\frac{1}{2^{2n}}\int_0^{\pi/2}{\sin^{2n+1}x\;dx}}$$

And here I am stuck, since I am not sure if I can do any change regarding the sum and integral, any help or tips is helpful. Thanks!

Answer 1

Your final expression has a small error. The equality you intended to write is

$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\sum_{n=0}^\infty \frac1{2^n}\int_0^{\pi/2}\sin^{2n+1}(x)\,dx$$

Now, if we change the order of the summation and integration (valid by uniform convergence), then we find that

$$\sum_{n=0}^\infty \frac{2^n}{(2n+1)\binom{2n}{n}}=\int_0^{\pi/2}\sin(x)\sum_{n=0}^\infty \frac1{2^n}\left(\sin^{2}(x)\right)^n\,dx$$

Next, sum the geometric series and carry out the resulting integral. Can you wrap this up now?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{n = 0}^{\infty}{2^{n} \over \pars{2n + 1}{2n \choose n}}} = \sum_{n = 0}^{\infty}2^{n}\,{\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 2}} \\[5mm] = &\ \sum_{n = 0}^{\infty}2^{n}\int_{0}^{1} x^{n}\pars{1 - x}^{n}\,\dd x = \int_{0}^{1}\sum_{n = 0}^{\infty} \bracks{2x\pars{1 - x}}^{\, n}\,\dd x \\[5mm] = &\ \int_{0}^{1}{\dd x \over 1 - 2x\pars{1 - x}} = {1 \over 2}\int_{0}^{1}{\dd x \over x^{2} - x + 1/2} \\[5mm] = &\ {1 \over 2}\int_{0}^{1}{\dd x \over \pars{x - 1/2}^{\, 2} + 1/4} = {1 \over 2}\int_{-1/2}^{1/2}{\dd x \over x^{2} + 1/4} \\[5mm] = &\ \int_{0}^{1/2}{\dd x \over x^{2} + 1/4} = 4\,{1 \over 2}\int_{0}^{1/2}{2\,\dd x \over \pars{2x}^{2} + 1} \\[5mm] = &\ 2\int_{0}^{1}{\dd x \over x^{2} + 1} = \bbx{\pi \over 2} \\ & \end{align}

Answer 3

Here is a short solution by a friend:

We have here

$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n{2n\choose n}}=\sum_{n=0}^\infty \frac{(2x)^{2n+1}}{(n+1){2n+2\choose n+1}}=\sum_{n=0}^\infty \frac{(2x)^{2n+1}}{(2n+1){2n\choose n}}$$

$$\overset{x=1/\sqrt{2}}{\Longrightarrow} \sum_{n=0}^\infty \frac{(\sqrt{2})^{2n+1}}{(2n+1){2n\choose n}}=\frac{\arcsin(\frac1{\sqrt{2}})}{\sqrt{1-1/2}}=\frac{\sqrt{2}\pi}{2}$$

$$\Longrightarrow \sum_{n=0}^\infty \frac{2^n}{(2n+1){2n\choose n}}=\frac{\pi}{2}$$

Answer 4

Start with beta function

$$\int_0^{\pi/2}\sin^{2a-1}(x)\cos^{2b-1}(x)dx=\frac12\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

Set $a=n+1$ and $b=1/2$ we have

$$\int_0^{\pi/2}\sin^{2n+1}(x)dx=\frac12\frac{\Gamma(n+1)\Gamma(1/2)}{\Gamma(n+3/2)}=\frac{\sqrt{\pi}}2\frac{\Gamma(n+1)}{(n+1/2)\Gamma(n+1/2)}$$

By Lengendre duplication formula $\Gamma(n+1/2)=\frac{\sqrt{\pi}\Gamma(2n+1)}{4^n\Gamma(n+1)}$ we get

$$\int_0^{\pi/2}\sin^{2n+1}(x)dx=\frac{4^n \Gamma^2(n+1)}{(2n+1)\Gamma(2n+1)}=\frac{4^n}{(2n+1){2n\choose n}}$$

Divide both sides by $2^n$ then $\sum_{n=0}^\infty$ we get

$$\sum_{n=0}^\infty\frac{2^n}{(2n+1){2n\choose n}}=\int_0^{\pi/2}\sin x\left(\sum_{n=0}^\infty\left(\frac{\sin^2x}{2}\right)^n\right)dx$$

$$=\int_0^{\pi/2}\sin x\left(\frac{1}{1-\frac{\sin^2x}{2}}\right)dx=\int_0^{\pi/2}\frac{2\sin x}{2-\sin^2x}dx$$

$$=\int_0^{\pi/2}\frac{2\sin x}{1+\cos^2x}dx=-2\arctan(\cos x)|_0^{\pi/2}=-2[\arctan(0)-\arctan(1)]=\frac{\pi}{2}$$