Proof without words for $\sum_{i=0}^\infty(-1)^i\frac{1}{2i+1}$

Question

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

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

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.

Answer 1

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!

Answer 2

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 $x∈[0;2π]$

Answer 3

$$|x|<1\implies \frac1{1+x^2}=1-x^2+x^4-x^6+\ldots\implies$$

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

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