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

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. $\square$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\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 ?}}$

\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}
A: 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
$$\begin{align}
  & =\sum\nolimits_{n=0}^{\infty }{\left\{ {{\left( -1 \right)}^{n}}\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{1}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}dx} \right\}} \\ 
 & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\sum\nolimits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\left( 1+{{x}^{2}} \right)}^{\frac{n}{2}+1}}}}dx} \\ 
 & =\frac{2}{\sqrt{\pi }}\int_{0}^{\infty }{\frac{dx}{\sqrt{1+{{x}^{2}}}\left( 1+\sqrt{1+{{x}^{2}}} \right)}} \\ 
 & =\frac{2}{\sqrt{\pi }}\left. \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right|_{0}^{\infty } \\ 
 & =\frac{2}{\sqrt{\pi }} \\ 
\end{align}$$
A: 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?
