Infinite Sum Including The Gamma Function How I can find the following the sum:
$$\sum_{n=1}^{+\infty}\frac{{{a}^{1-2n}}\ \Gamma \left( n-\frac{1}{2} \right)}{\Gamma \left( n \right)},\qquad a>0$$
Note:I have used mathematica which gives an exact result, namely 
 $\sqrt{\frac{\pi }{{{a}^{2}}-1}}$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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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With $\ds{a \in \mathbb{R}\setminus\bracks{-1,1}}$:

\begin{align}
\sum_{n = 1}^{\infty}a^{1 - 2n}\,
{\Gamma\pars{n - 1/2} \over \Gamma\pars{n}} & =
\Gamma\pars{1 \over 2}\sum_{n = 1}^{\infty}a^{1 - 2n}\,
{\pars{n - 3/2}! \over \pars{n - 1}!\pars{-1/2}!} =
\root{\pi}\sum_{n = 1}^{\infty}a^{1 - 2n}{n - 3/2 \choose n - 1}
\\[5mm] & =
\root{\pi}\sum_{n = 0}^{\infty}a^{-1 - 2n}{n - 1/2 \choose n} =
\root{\pi}\sum_{n = 0}^{\infty}a^{-1 - 2n}{-1/2 \choose n}\pars{-1}^{n}
\\[5mm] & =
{\root{\pi} \over a}\sum_{n = 0}^{\infty}{-1/2 \choose n}
\pars{-\,{1 \over a^{2}}}^{n} =
{\root{\pi} \over a}\bracks{1 + \pars{-\,{1 \over a^{2}}}}^{-1/2}
\label{1}\tag{1}
\\[5mm] & =
\bbx{\mrm{sgn}\pars{a}\root{\pi \over a^{2} - 1}}
\end{align}
A: Idea: use the Legendre duplication formula
$$\Gamma(z) \; \Gamma\left(z + \frac12\right) = 2^{1 - 2z} \; \sqrt{\pi} \; \Gamma(2z).$$
Then,
$$
\Gamma(n - 1/2) =
\frac{2^{2 - 2n} \; \sqrt{\pi} \; \Gamma(2n - 1)}{\Gamma(n)}
$$
