# 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 '13 at 4:07
@tomasz Could you convert your comment into an answer so that this question gets answered? –  user17762 May 15 '13 at 4:25

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

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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.

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