Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
8  
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
add comment

3 Answers 3

up vote 4 down vote accepted

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\root{2}= \lim_{n \to \infty}\sum_{i = 0}^{n}{\pars{-1}^{i}\pars{-\,\half}_{i} \over i!}:\ {\Large ?}}$

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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.