# 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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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}

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