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$\ds{\raíz{2}=
\lim_{n \to \infty}\sum_{i = 0}^{n}{\pars{-1}^{i}\pars{-\,\mitad}_{i} \over i!}:\ {\Large ?}}$
Tenga en cuenta que
\begin{align}
\pars{-\,\half}_{i}&=\pars{-1}^{i}\pars{\half - i + 1}_{i}
=\pars{-1}^{i}\,{\Gamma\pars{3/2} \over \Gamma\pars{3/2 - i}}
\\[3mm]&=\pars{-1}^{i}\,\Gamma\pars{3 \over 2}\,
{\Gamma\pars{-1/2 + i}\sin\pars{\pi\bracks{-1/2 + i}} \over \pi}
=-\,{1 \over 2\root{\pi}}\,\Gamma\pars{-\,\half + i}
\\[3mm]&=-\,{1 \over 2\root{\pi}}\,\int_{0}^{\infty}t^{-3/2 + i}\expo{-t}\,\dd t\,,
\qquad i \geq 1\quad\mbox{and}\quad\pars{-\,\half}_{0} = 1
\end{align}
\begin{align}
\color{#00f}{\large\sum_{i = 0}^{\infty}{\pars{-1}^{i}\pars{-\,\half}_{i} \over i!}}
&=
1 + \sum_{i = 1}^{\infty}{\pars{-1}^{i}\over i!}\bracks{-\,{1 \over 2\root{\pi}}\int_{0}^{\infty}t^{-3/2 + i}\expo{-t}\,\dd t}
\\[3mm]&=1 - {1 \over 2\root{\pi}}\int_{0}^{\infty}t^{-3/2}\expo{-t}
\sum_{i = 1}^{n}{\pars{-1}^{i}t^{i}\over i!}\,\dd t
\\[3mm]&=1 -\,{1 \over 2\root{\pi}}\int_{0}^{\infty}t^{-3/2}\expo{-t}
\pars{\expo{-t} - 1}\,\dd t
\\[3mm]&
=1 - {1 \over 2\root{\pi}}\
\overbrace{\int_{0}^{\infty}t^{-3/2}\pars{\expo{-2t} - \expo{-t}}\,\dd t}
^{\ds{2\pars{1 - \root{2}}\pi}}
=\color{#00f}{\large\root{2}}
\end{align}
\begin{align}
&\int_{0}^{\infty}t^{-3/2}\pars{\expo{-2t} - \expo{-t}}\,\dd t
=\int_{t = 0}^{t \to \infty}\pars{\expo{-2t} - \expo{-t}}\,\dd\pars{-2t^{-1/2}}
\\[3mm]&=-\int_{0}^{\infty}\pars{-2t^{-1/2}}\pars{-2\expo{-2t} + \expo{-t}}\,\dd t
=-4\int_{0}^{\infty}t^{-1/2}\expo{-2t}\,\dd t
+ 2\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t
\\[3mm]&=-2\root{2}\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t
+ 2\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t
=2\pars{1 - \root{2}}\ \underbrace{\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t}
_{\ds{=\ \Gamma\pars{\half}\ =\ \root{\pi}}}
\end{align}