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.

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., 
$$
\sqrt{2}=\sqrt{1+1}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}.
$$
