TITLE: How to compute a $\sum_{n=0}^{\infty}{\frac{2^n}{(2n+1){2n\choose n}}}$
QUESTION [8 upvotes]: 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!

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}