Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the follow up gives me $\infty$. How to approach this type of exercises?
The result should be $1/2$
 A: HINT:
Note that
$$\begin{align}
\frac{n}{(2n+1)!!}&=\frac12\left( \frac{2n+1-1}{(2n+1)!!}\right)\\\\&=\frac12\left(\frac{1}{(2n-1)!!}-\frac1{(2n+1)!!}\right)
\end{align}$$
Now, telescope.
A: Observe that
$$
\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot3\cdot 5\cdots (2n+1)}\right)
$$
Hence
$$\sum_{n=1}^\infty\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\sum_{n=1}^\infty\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot 3\cdot 5\cdots (2n+1)}\right)=\frac{1}{2}\left(1-\frac{1}{1\cdot3}+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot 5}-\frac{1}{1\cdot3\cdot 5}+\cdots\right)=\frac{1}{2}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{n \over 1 \times 3 \times 5 \times \pars{2n + 1}}} =
\sum_{n = 1}^{\infty}{n \over \prod_{k = 0}^{n}\pars{2k + 1}} =
\sum_{n = 1}^{\infty}{n \over 2^{n + 1}\prod_{k = 0}^{n}\pars{k + 1/2}}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}{n \over 2^{n}}\,
{1 \over \pars{1/2}^{\overline{n + 1}}} =
{1 \over 2}\sum_{n = 1}^{\infty}{n \over 2^{n}}\,
{1 \over \Gamma\pars{1/2 + n + 1}/\Gamma\pars{1/2}}
\\[5mm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}{n \over 2^{n}}\,{1 \over n!}\,
{\Gamma\pars{n + 1}\Gamma\pars{1/2} \over \Gamma\pars{n + 3/2}} =
{1 \over 4}\sum_{n = 0}^{\infty}{1 \over 2^{n}\, n!}\,
{\Gamma\pars{n + 2}\Gamma\pars{1/2} \over \Gamma\pars{n + 5/2}}
\\[5mm] = &\
{1 \over 4}\sum_{n = 0}^{\infty}{1 \over 2^{n}\, n!}\,
\int_{0}^{1}t^{n + 1}\pars{1 - t}^{-1/2}\,\dd t =
{1 \over 4}\int_{0}^{1}{t \over \root{1 - t}}\sum_{n = 0}^{\infty}{\pars{t/2}^{n} \over n!}\,\dd t
\\[5mm] = &\
{1 \over 4}\int_{0}^{1}{t\expo{t/2} \over \root{1 - t}}\,\dd t =
{1 \over 4}\pars{\vphantom{\Large A}-2\expo{t/2}\root{1 - t}}_{\ 0}^{\ 1} = \bbx{1 \over 2}
\end{align}
