<s>[INST] How does one show that $\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{\Gamma\left({2n+1\over 2}\right)\over n!}=\sqrt{{x\pi\over x+1}}?$ Consider $(1)$

$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{\Gamma\left({2n+1\over 2}\right)\over n!}=S\tag1$$
  $x>0$
How does one show that $S=\sqrt{{x\pi\over x+1}}$

An attempt:
$$\Gamma\left({1\over 2}+n\right)={(2n-1)!!\sqrt{\pi}\over 2^n}$$
$(1)$ becomes
$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{(2n-1)!!\over 2^n}\cdot{1\over n!}=\sqrt{x\over x+1}\tag2$$
further to
$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{1\over 2^n(2n-2)!!}=\sqrt{x\over x+1}\tag3$$
 [/INST] We note that
$$\frac{\Gamma \left ( n+\frac12 \right )}{\Gamma \left ( n+1 \right )} = \frac1{\sqrt{\pi}} \int_0^1 dy \, y^{-1/2} (1-y)^{n-1/2} $$
Then
$$\begin{align}\sum_{n=0}^{\infty} \frac{\Gamma \left ( n+\frac12 \right )}{\Gamma \left ( n+1 \right )} w^n &= \frac1{\sqrt{\pi}} \int_0^1 dy \, y^{-1/2} (1-y)^{-1/2} \sum_{n=0}^{\infty} [(1-y) w]^n \\ &= \frac1{\sqrt{\pi}} \int_0^1 dy \, \frac{y^{-1/2} (1-y)^{-1/2}}{1-w (1-y)} \\ &= \frac{2}{\sqrt{\pi}} \int_0^1 dy \frac{(1-y^2)^{-1/2}}{1-w (1-y^2)} \\ &= \frac1{2 \sqrt{\pi}} \int_0^{2 \pi} \frac{d\theta}{1-w \sin^2{\theta}} \end{align}$$
where we assume that $|w| \lt 1$.  That last integral may be performed any number of ways, e.g., extension to the complex plane and the residue theorem.  The result is $2 \pi/\sqrt{1-w}$.  Thus,
$$S = \sqrt{\frac{\pi}{1+\frac1{x}}} = \sqrt{\frac{\pi x}{x+1}} $$
ADDENDUM
Let's illustrate how to do that last integral.
$$\int_0^{2 \pi} \frac{d\theta}{1-w \sin^2{\theta}} = 2 \int_0^{\pi} \frac{d\theta}{1-w \sin^2{\theta}} =  2 \int_0^{\pi} \frac{d\theta}{1-w/2 + (w/2) \cos{2 \theta}} = \int_0^{2 \pi}\frac{d\theta}{1-w/2+(w/2) \cos{\theta}}$$
The above manipulations serve to show how to reduce a potentially messy problem into one a lot less messy.  In this case, we have reduced the number of poles in the complex plane from four to two.
Now express as a complex integral using $z=e^{i \theta}$; the integral is equal to
$$-i 2 \oint_{|z|=1} \frac{dz}{(w/2) z^2 + (2-w) z+(w/2)} $$
The poles of the integrand are at $z_{\pm}=[-(2-w) \pm 2 \sqrt{1-w}]/w$, and the only pole inside the unit circle is $z_+$.  Thus, by the residue theorem, the integral is equal to $i 2 \pi$ times the residue at that pole, the residue being equal to $1/(2 \sqrt{1-w})$.  Thus the integral is equal to $2 \pi/\sqrt{1-w}$ as asserted.
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