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

Question

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

$$\sum_{n=0}^{\infty}\left(-{1\over x}\right)^n\cdot{1\over 2^n(2n-2)!!}=\sqrt{x\over x+1}\tag3$$

Answer 1

We wish to calculate ($x>1$)

$$ S=\sum_{n\geq0}\frac{(-1/x)^n}{n!}\Gamma(n+1/2) $$

we have that $\Gamma(n+1/2)=\int_0^{\infty} t^{n-1/2}e^{-t}dt$

so

$$ S=\int_0^{\infty}dte^{-t}t^{-1/2}\sum_{n\geq 0}\frac{(-t/x)^n}{n!}=\int_0^{\infty}dte^{-t(1+1/x)}t^{-1/2}=\frac{\sqrt{\pi x}}{\sqrt{1+x}} $$

the interchange of sum and integral is justified since $\int_0^{\infty}e^{-t}t^z$ converges absolutly for any $z>-1$ and the last integral can be turned into a simple Gaussian one via $t\rightarrow t^2$

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note the non obvious limit $\lim_{x\rightarrow\infty} S(x)=\sqrt{\pi}$

Answer 2

$$ \sum_{n=0}^\infty \left(-\frac{1}{x}\right)^n\frac{1}{n!}\frac{1}{4^n}\frac{(2n)!}{n!}\sqrt{\pi} = \sum_{n=0}^\infty \left(-\frac{1}{4x}\right)^n\frac{(2n)!}{(n!)^2}\sqrt{\pi} = \sum_{n=0}^\infty \left(-\frac{1}{4x}\right)^n\left(\matrix{2n\\n}\right)\sqrt{\pi} $$ the final is of the form $$ \sum_{k=0}^\infty \left(\matrix{2k\\k}\right)z^k = \frac{1}{\sqrt{1-4z}}\;\;\text{iff}\;\; |z| <1/4 $$ (see wiki)

if we have $$ \left|\frac{1}{-4x}\right| =\left|\frac{1}{4x}\right| < \frac{1}{4} $$ then we have $$ \sum_{n=0}^\infty \left(-\frac{1}{4x}\right)^n\left(\matrix{2n\\n}\right)\sqrt{\pi} = \frac{1}{\sqrt{1-4\frac{1}{-4x}}}\sqrt{\pi}= \sqrt{\frac{x\pi}{1+x}} $$

Answer 3

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.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\sum_{n = 0}^{\infty}\pars{-\,{1 \over x}}^{n}\, {\Gamma\pars{\bracks{2n + 1}/2}\over n!} = \pars{-\,{1 \over 2}}!\sum_{n = 0}^{\infty} {n - 1/2 \choose n}\pars{-\,{1 \over x}}^{n} \\[5mm] = &\ \root{\pi}\sum_{n = 0}^{\infty}\underbrace{\bracks{{- 1/2 \choose n}\pars{-1}^{n}}}_{\ds{\underbrace{=\ {n - 1/2 \choose n}}_{\ds{\mbox{Negating Property}}}}}\ \pars{-\,{1 \over x}}^{n} = \root{\pi}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{1 \over x}^{n} \\[5mm] = &\ \root{\pi}\pars{1 + {1 \over x}}^{-1/2} = \bbx{\ds{\root{\pi x \over x + 1}}}\,,\qquad\qquad \verts{x} > 1. \end{align}

Answer 5

We may also prove this by working backwards from the integral $\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ and applying Ramanujan's master theorem.

We start with the following integral: $$\begin{align*} \int_0^\infty\sqrt{\frac{\pi/t}{1+1/t}}t^{s-1}dt &=\sqrt{\pi}\int_0^\infty\frac{u^{-s-1/2}}{(1+u)^{1/2}}du,\;\;t=u^{-1} \\ &=\sqrt{\pi}\int_0^\infty\frac{u^{1/2-s-1}}{(1+u)^{1/2-s+s}}du \\ &=\Gamma(1/2-s)\Gamma(s),\;\;0<\Re [s]<1/2. \end{align*}$$

Which implies that $\phi(s)=\Gamma(1/2+s)$. So we then have

$$\begin{align*}\sqrt{\frac{\pi/t}{1+1/t}}&=\sum_{n=0}^\infty\frac{\phi(n)}{n!}(-t)^n\\ &=\sum_{n=0}^\infty\frac{\Gamma(1/2+n)}{n!}(-t)^n,\;\;|t|<1. \end{align*} $$

Replacing $t$ with $1/t$ yields your sum.