Demonstrate: $ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $ Demonstrate that $\sqrt2$ can be expressed as:
$$ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $$ Where $\left(z\right)_i$ is the Pochhammer symbol
$\left(z\right)_i=z(z+1)(z+2)...(z+i-1);  (z)_0=1$
This is a nice problem, just wanted to share it.
 A: If we Taylor expand $f(x)=\sqrt{1+x}$, around $x=0$, we get
$$
\sqrt{1+x}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}x^k=\sum_{k=0}^\infty
a_kx^k, \tag{1}
$$
and this series converges for $|x|<1$, as
$$
|a_k|=\frac{1}{k!}\cdot\frac{1}{2}\left(\frac{1}{2}\frac{3}{2}\cdots\frac{k-1-\frac{1}{2}}{2}\right)=\frac{1}{2k}\prod_{j=1}^{k-1}\frac{j-1/2}{j}<1.
$$
Also
$$
\left|\frac{a_{k+1}}{a_{k}}\right|=\frac{k}{k+1}\cdot
\frac{k-\frac{1}{2}}{k}=\frac{k-\frac{1}{2}}{k+1}=1-\frac{3}{2(k+1)},
$$
and hence
$$
\lim_{k\to\infty}k\left(\left|\frac{a_{k+1}}{a_{k}}\right|-1\right)=-\frac{3}{2.}
$$
which, due to Raabe's test, converges absolutely, even for $|x|=1$,  and therefore, the powerseries $(1)$ defines a continuous function for $x\in [-1,1]$. Thus, $(1)$ holds even for $x=1$, i.e., 
$$
\sqrt{2}=\sqrt{1+1}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}.
$$
A: By generalizing Newton's Binomial Theorem to non-natural exponents, we get the Binomial Series, which, for $n=\frac12$ , yields the following result :
$$\sum_{k=0}^nC_n^k=2^n\qquad\iff\qquad\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n)!!}=\sqrt2$$
Generalizing Vandermonde's Identity to non-natural arguments like $n=\frac12$ , and using Particular Values of the Gamma Function, we deduce the following identity :
$$\sum_{k=0}^n\left(C_n^k\right)^2=C_{2n}^n\qquad\iff\qquad\sum_{n=0}^\infty\left[\frac{(2n-3)!!}{(2n)!!}\right]^2=\frac4\pi$$ where !! represents the double factorial.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\root{2}=
     \lim_{n \to \infty}\sum_{i = 0}^{n}{\pars{-1}^{i}\pars{-\,\half}_{i} \over i!}:\                 {\Large ?}}$

Note that
  \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}

