Infinite Sum Including The Gamma Function

Question

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 $\sqrt{\frac{\pi }{{{a}^{2}}-1}}$

Answer 1

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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}

Answer 2

Idea: use the Legendre duplication formula $$\Gamma(z) \; \Gamma\left(z + \frac12\right) = 2^{1 - 2z} \; \sqrt{\pi} \; \Gamma(2z).$$ Then, $$ \Gamma(n - 1/2) = \frac{2^{2 - 2n} \; \sqrt{\pi} \; \Gamma(2n - 1)}{\Gamma(n)} $$