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$$\sum_{i=0}^\infty(-1)^i\frac{1}{2i+1}$$

$$1-\frac13+\frac15-\frac17+\frac19-\cdots=\frac\pi4$$

Does anyone know of a proof without words for this? I am not looking a for a just any proof, since I can prove it myself. What I am looking for is a elegant physical interpretation, or anything else that matches that kind of beauty.

Just to clear up, the proof I already know is

$$\int_0^1\left(1-x^2+x^4-x^6+x^8-\cdots\right)\,\mathrm{d}x = \int_0^1\frac{\mathrm{d}x}{1+x^2}$$

So I am intersted in anything that is not obviously related to this.

Hawk
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Guy
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3 Answers3

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I believe it may be possible to show this formula using Fourier series expansion of the following function $f$ periodic with period $2\pi$ such that $f(x)=\frac{π−x}{2} $ for $x∈[0;2π]$

Jack
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  • I understand your function, how would it help? – Guy Mar 16 '14 at 11:43
  • I believe (to justify we use Dirichlet's theorem) that the fourier expansion of $f$ is $f(x)=\sum \limits_{k=1}^∞\frac{sin(kx)}{k}$, if we evaluate this function at $\frac{\pi}{2}$ ( a point where $f$ is continuous) we find your formula. – Jack Mar 16 '14 at 11:47
  • ah. very nice. +1 – Guy Mar 16 '14 at 11:48
  • I cannot accept though. I want proof without words. – Guy Mar 16 '14 at 11:49
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$$|x|<1\implies \frac1{1+x^2}=1-x^2+x^4-x^6+\ldots\implies$$

$$\arctan x=\int\frac{dx}{1+x^2}=x-\frac13x^3+\frac15x^5-\frac17x^7+\ldots\;,\;\;\forall\,x\in(-1,1\color{red}]\;(<-- !!)\implies$$

$$\frac\pi4=\arctan 1=1-\frac13+\frac15-\frac17+\ldots$$

DonAntonio
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  • I think you misunderstood, I wanted a proof without words(it might not be possible, just on the search for one). I already know this as stated at the end. – Guy Mar 16 '14 at 13:46
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    Yup, but you began with a rather non intuitive equality between an integral of infinite summands. I first show the power series of $;\frac1{1+x^2};$, followed by the fact that one can integrate such a series at least within the convergence radius, getting a series with an extra point of convergence ($;x=1;$) which, finally and upon substitution of $;x;$ by this point, gives you the desired equality. – DonAntonio Mar 16 '14 at 13:49
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    That's very well, I just stated it so that this answer wouldn't arise. (meaning that I am looking for alternatives,that equation was just a handwaving that I know this already). Btw, the downvote isn't mine. I'll upvote this to keep it at 0 at least. – Guy Mar 16 '14 at 13:56
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expand arctan(x) into taylor series: $$ \arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots $$ take x=1 and you get what you want!

Robert Fan
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