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

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

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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}