Find sum of Series, Binomial Theorem how would you find the sum of the series
$\sum_{n=1}^{\infty} \frac{\pi(\pi-1)...(\pi-n+1)}{2^nn!}$ I know that it relates to the binomial theorem but I'm having trouble finding thhe function that the series is derived for
 A: $\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}}}
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\begin{align}
\sum_{n = 1}^{\infty}
{\pi\pars{\pi - 1}\ldots\pars{\pi - n + 1} \over 2^{n}\, n!} & =
\sum_{n = 1}^{\infty}
{\pi! \over  n!\pars{\pi - n}!}\pars{1 \over 2}^{n} =
\sum_{n = 1}^{\infty}
{\pi \choose n}\pars{1 \over 2}^{n}
\\[5mm] & =
\bbx{\pars{1 + {1 \over 2}}^{\pi} - 1 = \pars{3 \over 2}^{\pi} - 1}
\approx 2.5744
\end{align}
A: We have $(u+1)^{\pi}=1+\displaystyle\sum_{n=1}^{\infty}\dfrac{\pi(\pi-1)\cdots(\pi-n+1)}{n!}u^{n}$. Now put $u=x/2$ to conclude.
