Is $\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}{2})}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}$ true?

Question

Prove/disprove $$\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\frac{2}{\sqrt{\pi}}.\tag 1$$

As far as I can tell, this is true, although it seems to converge very slowly.

I came up with a proof but I don't know if it's valid.

Let $$J=\int_0^\pi \frac{xdx}{1+\sin x}.$$ On one hand, we have $$\frac1{1+\sin x}=\sum_{n\ge0}(-1)^n\sin(x)^n,$$ so that $$J=\sum_{n\ge0}(-1)^n p_n,\tag 2$$ where $$ \begin{align} p_n&=\int_0^\pi x\sin(x)^ndx\\ &=\int_\pi^0 -(\pi-x)\sin(\pi-x)^ndx\\ &=\pi\int_0^\pi\sin(x)^ndx-p_n\\ \Rightarrow p_n&=\frac\pi2\int_0^\pi\sin(x)^ndx. \end{align} $$ And since $\sin(x)=\sin(\pi-x)$, $$p_n=\pi\int_0^{\pi/2}\sin(x)^ndx=\frac{\pi^{3/2}}{2}\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}.\tag 3$$

On the other hand, we have $1+\sin x=2\sin^2(\tfrac{x}2-\tfrac\pi4)$, so that $$\begin{align} J&=\frac12\int_0^\pi\frac{xdx}{\sin^2(\tfrac{x}2-\tfrac\pi4)}\\ &=2\int_{\pi/4}^{3\pi/4}\frac{tdt}{\sin^2t}-\frac\pi2\int_{\pi/4}^{3\pi/4}\frac{dt}{\sin^2 t}\\ &=2\left(\ln\sin x-x\cot x\right)\bigg|_{\pi/4}^{3\pi/4}-\frac\pi2\left(-\cot x\right)\bigg|_{\pi/4}^{3\pi/4}\\ &=2\pi-\frac\pi2\cdot2=\pi. \end{align}$$ Then from $(2)$ and $(3)$, we have $$\frac{\pi^{3/2}}{2}\sum_{n\ge0}(-1)^n\frac{\Gamma(\tfrac{n+1}2)}{\Gamma(\tfrac{n}2+1)}=\pi,$$ which is equivalent to $(1)$. $\square$

Can you come up with any other proofs to $(1)$? Thanks!

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Edit (11/12/2020):

Here is a proof that the interchange of the sum and integral in $(2)$ is valid.

The partial sums $$S_M(x)=\sum_{n=0}^M(-1)^n\sin(x)^n$$ form a uniformly convergent sequence of functions for $x$ in $[0,\pi/2)$ or $(\pi/2,\pi]$, and they converge to the limit $$\lim_{M\to\infty}S_M(x)=\frac1{1+\sin x},\qquad x\in[0,\pi]\setminus\{\pi/2\}.$$ Choose $\varepsilon>0$ and notice that $$J=\int_{0}^{\pi}\frac{xdx}{1+\sin x}=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\int_0^{\pi/2-\varepsilon}\frac{xdx}{1+\sin x}+\int_{\pi/2+\varepsilon}^\pi\frac{xdx}{1+\sin x}.$$ The sums $S_M(x)$ converge uniformly as $M\to\infty$ when $x\in[0,\pi/2-\varepsilon]\cup[\pi/2+\varepsilon,\pi]$, so we can interchange the sum and integral to get $$J=\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}+\sum_{n\ge0}(-1)^n(a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)),$$ where $$\begin{align} a_n(t)&=\int_0^t x\sin(x)^ndx\\ b_n(t)&=\int_t^\pi x\sin(x)^ndx. \end{align}$$ We have $a_n(t)+b_n(t)=p_n$ for all $t\in[0,\pi]$. As $\varepsilon$ approaches $0$, $\int_{\pi/2-\varepsilon}^{\pi/2+\varepsilon}\frac{xdx}{1+\sin x}$ approaches $0$.

And since $a_n(t), b_n(t)$ are continuous, $a_n(\pi/2-\varepsilon)+b_n(\pi/2+\varepsilon)$ approaches $a_n(\pi/2)+b_n(\pi/2)=p_n$, so that $$J=\sum_{n\ge0}(-1)^np_n$$ as desired. $\square$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(\bracks{n+1}/2) \over \Gamma(n/2 + 1)} = {2 \over \root{\pi}}} \approx 1.1284:\ {\Large ?}}$

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\begin{align} &\bbox[5px,#ffd]{\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(\bracks{n+1}/2) \over \Gamma(n/2 + 1)}} \\[5mm] = &\ {1 \over \root{\pi}}\sum_{n\ \geq\ 0}\pars{-1}^{n}\,{\Gamma(n/2 + 1/2)\Gamma\pars{1/2} \over \Gamma(n/2 + 1)} \\[5mm] = &\ {1 \over \root{\pi}}\sum_{n\ \geq\ 0}\pars{-1}^{n} \int_{0}^{1}t^{n/2 - 1/2}\,\,\, \pars{1 - t}^{-1/2}\,\,\dd t \\[5mm] = &\ {1 \over \root{\pi}}\int_{0}^{1}{1 \over \root{t}\root{1 - t}}\sum_{n\ \geq\ 0}\pars{-\root{t}}^{n}\,\dd t \\[5mm] = &\ {1 \over \root{\pi}}\int_{0}^{1}{1 \over \root{t}\root{1 - t}}{1 \over 1 + \root{t}}\,\dd t \\[5mm] \stackrel{t\ \mapsto\ t^{2}}{=} &\ {2 \over \root{\pi}}\int_{0}^{1}{1 \over \root{1 - t^{2}}}{1 \over 1 + t}\,\dd t \\[5mm] \stackrel{t\ \mapsto\ \sin\pars{\theta}}{=} &\ {2 \over \root{\pi}}\int_{0}^{\pi/2} {\dd\theta \over 1 + \sin\pars{\theta}} \\[5mm] = &\ {2 \over \root{\pi}}\int_{0}^{\pi/2} \bracks{\sec^{2}\pars{\theta} - \sec\pars{\theta}\tan\pars{\theta}}\dd\theta \\[5mm] = &\ {2 \over \root{\pi}}\ \underbrace{\bracks{\sin\pars{\theta} - 1 \over \cos\pars{\theta}}_{0}^{\pars{\pi/2}^{\,-}}} _{\ds{=\ 1}}\ =\ \bbx{2 \over \root{\pi}} \approx 1.1284 \\ & \end{align}

Answer 2

Based on this answer: $$\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}dx}=\frac{\Gamma \left( \frac{n+1}{2} \right)}{\Gamma \left( \frac{n}{2}+1 \right)}$$ so the sum in question $$\begin{align} & =\sum\nolimits_{n=0}^{\infty }{\left\{ {{\left( -1 \right)}^{n}}\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}dx} \right\}} \\ & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\sum\nolimits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}}dx} \\ & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{dx}{\sqrt{1+{{x}^{2}}}\left( 1+\sqrt{1+{{x}^{2}}} \right)}} \\ & =\frac{2}{\sqrt{\pi }}\left. \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right|_{0}^{\infty } \\ & =\frac{2}{\sqrt{\pi }} \\ \end{align}$$

Answer 3

Using the well-known Maclaurin expansions for $\arcsin$ and its square, we first obtain

$$\begin{align*} \arcsin x &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n} (2n+1)} x^{2n+1} \\ \implies \frac1{\sqrt{1-x^2}} &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n}} x^{2n} \color{red}{=: F(x)} \\[2ex] \arcsin^2x &= \sum_{n=1}^\infty \frac{2^{2n-1}}{n^2 \binom{2n}n} x^{2n} \\ \implies \frac{2x\arcsin x}{\sqrt{1-x^2}} &= \sum_{n=1}^\infty \frac{2^{2n}}{n \binom{2n}n} x^{2n} \\ \implies \frac{2x^2}{1-x^2} + \frac{2x\arcsin x}{\left(1-x^2\right)^{3/2}} &= \sum_{n=1}^\infty \frac{2^{2n+1}}{\binom{2n}n} x^{2n} \\ \implies \frac{2\arcsin x}{\sqrt{1-x^2}} &= \sum_{n=0}^\infty \frac{2^{2n+1}}{(2n+1) \binom{2n}n} x^{2n+1} \color{blue}{=: G(x)} \end{align*}$$

Now in the sum of interest, convert from gamma to binomial coefficient, then the sum to a limit.

$$\begin{align*} & \sum_{n\ge0} (-1)^n \frac{\Gamma\left(\frac n2+\frac12\right)}{\Gamma\left(\frac n2 + 1\right)} \\ &= \sum_{n\ge0} \left[\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(n + 1\right)} - \frac{\Gamma\left(n+1\right)}{\Gamma\left(n+\frac32\right)}\right] \\ &= \sum_{n\ge0} \left[\Gamma\left(\frac12\right) \frac{\binom{2n}n}{2^{2n}} - \frac1{\Gamma\left(\frac12\right)} \frac{2^{2n+1}}{(2n+1) \binom{2n}n}\right] \\ &= \frac1{\sqrt\pi} \lim_{x\to1^-} \left[\pi \color{red}{F(x)} - \color{blue}{G(x)}\right] = \boxed{\frac2{\sqrt\pi}} \end{align*}$$

Answer 4

Under the Legendre duplication formula, $$\Gamma(z)\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$$ making: $$\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{\Gamma(z)}$$ and so: $$\frac{\Gamma(z+\frac12)}{\Gamma(z+1)}=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{\Gamma(z)\Gamma(z+1)}=2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{z\Gamma^2(z)}$$ could you use this?