Find sum of Series, Binomial Theorem

Question

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

Answer 1

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

Answer 2

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.