# 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.

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Isn't it just a special case of the rule $(1+a)^\alpha=\sum_{i=0}^\infty {\alpha \choose i} a^i$? –  tomasz Jan 2 at 4:07
@tomasz Could you convert your comment into an answer so that this question gets answered? –  user17762 May 15 at 4:25
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.