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I am hoping to find out where the formula $$\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}$$ comes from. I can't see how one could begin to prove it.

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marked as duplicate by Simply Beautiful Art, Lord Shark the Unknown, Nosrati, Namaste, JonMark Perry Dec 17 '17 at 5:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ math.stackexchange.com/a/14116/42969 $\endgroup$ – Martin R Feb 17 '17 at 22:04
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    $\begingroup$ If you are wondering on the untimely fast response, it is because I had derived this formula just last month! :D AH! I am so happy to write this answer. $\endgroup$ – Simply Beautiful Art Feb 17 '17 at 22:19
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    $\begingroup$ And I am so happy to find people like you here :) I wish anything I paid for was delivered in an 'untimely fast' manner. Passion is such an awesome motivator. $\endgroup$ – J.A.K. Feb 18 '17 at 1:45
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    $\begingroup$ $$\sum_{k=0}^{\infty}\frac{k!}{\left(2k+1\right)!!}x^k=\frac{2 \sin ^{-1}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)}{\sqrt{(2-x) x}}$$ $\endgroup$ – Claude Leibovici Feb 18 '17 at 6:40
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    $\begingroup$ @SimplyBeautifulArt. You are totally correct ! I found the result (from a CAS) really nice. Cheers. $\endgroup$ – Claude Leibovici Feb 18 '17 at 12:27
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Let us start with the geometric series:

$$\frac1{1-r}=\sum_{k=0}^\infty r^k$$

If we let $r=-x^2$ and integrate both sides from zero to one, we get the famous Leibniz formula for $\pi$.

$$\frac\pi4=\arctan(1)=\int_0^1\frac1{1+x^2}\ dx=\int_0^1\sum_{k=0}^\infty(-x^2)^k\ dx=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$

Applying an Euler Transform, we arrive at

$$\begin{align}\frac\pi4&=\sum_{n=0}^\infty\frac1{2^{1+n}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{2k+1}\\\\&=\sum_{n=0}^\infty\text{simplifying the inner sum}\\\\&=\frac12\sum_{n=0}^\infty\frac{k!}{(2k+1)!!}\end{align}$$

The simplifying step comes by noting that

$$\sum_{k=0}^0\binom0k\frac{(-1)^k}{2k+1}=\frac11=2^0\frac{0!}{1!!}\color{green}\checkmark$$

$$\sum_{k=0}^1\binom1k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)=2^1\frac{1!}{3!!}\color{green}\checkmark$$

$$\sum_{k=0}^2\binom2k\frac{(-1)^k}{2k+1}=\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)=2^2\frac{2!}{5!!}\color{green}\checkmark$$

$$\sum_{k=0}^3\binom3k\frac{(-1)^k}{2k+1}=\left[\left(\frac11-\frac13\right)-\left(\frac13-\frac15\right)\right]-\left[\left(\frac13-\frac15\right)-\left(\frac15-\frac17\right)\right]=2^3\frac{3!}{7!!}\color{green}\checkmark$$

You can prove by induction (and some observation) that the denominators are clearly odd double factorials, and with some work, you can derive the numerators.

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    $\begingroup$ Yes, this is one of my favourite posts as well. $(+1)$ :D $\endgroup$ – Mr Pie Oct 6 '18 at 17:36
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's well known that $\ds{\pars{2k + 1}!! = {\pars{2k + 2}! \over 2^{k + 1}\pars{k + 1}!}}$ such that

\begin{align} \sum_{k = 0}^{\infty}{k! \over \pars{2k + 1}!!} & = \sum_{k = 0}^{\infty}{k!\pars{k + 1}! \over \pars{2k + 2}!}\,2^{k + 1} = \sum_{k = 0}^{\infty}{\Gamma\pars{k + 1}\Gamma\pars{k + 2} \over \Gamma\pars{2k + 3}}\,2^{k + 1} \\[5mm] & = \sum_{k = 0}^{\infty}2^{k + 1}\int_{0}^{1}x^{k}\pars{1 - x}^{k + 1}\,\dd x = 2\int_{0}^{1}\pars{1 - x}\sum_{k = 0}^{\infty}\bracks{2x\pars{1 - x}}^{\,k} \,\dd x \\[5mm] & = 2\int_{0}^{1}\pars{1 - x}{1 \over 1 - 2x\pars{1 - x}}\,\dd x = \int_{0}^{1}{1 - x \over x^{2} - x + 1/2}\,\dd x = \int_{-1/2}^{1/2}{1/2 - x \over x^{2} + 1/4}\,\dd x \\[5mm] & = 2\int_{0}^{1}{\dd x \over x^{2} + 1} = 2\arctan\pars{1} = 2\,{\pi \over 4} = \bbx{\ds{\pi \over 2}} \end{align}

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  • $\begingroup$ Wonderful! +1 I was wondering how to start with the series and snake my way back to the $\pi$, but failed to see the beautiful Beta function. Might want to reference that part for the viewers. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 2:48
  • $\begingroup$ @SimplyBeautifulArt Thanks. Beta is always a nice 'killer'. $\endgroup$ – Felix Marin Feb 18 '17 at 2:49
  • $\begingroup$ Lol, I'm still looking at it in awe, as those factorials indeed worked out perfectly. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 2:52

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