It is well known that the alternate sum of the reciprocals of the odd numbers adds up to $\frac{\pi}{4}$. That is $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots = \sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac{\pi}{4}.$$ A proof can be obtained by using geometric series and termwise integration to compute the Taylor series of $\arctan(x)$ and then evaluating it at $x=1$: $$\arctan'(x)=\frac{1}{1+x^2}=\sum_{k=0}^\infty(-1)^kx^{2k}\Rightarrow\arctan(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{2k+1}\Rightarrow\arctan(1)= \sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$ and now use that $\arctan(1)=\frac{\pi}{4}$.
Now, is there any visual (geometric) or other sort of intuitive proof on why this should be true?
An example of the sort of visual proof I'm thinking would be something like starting by taking a quarter circle inscribed in an area 1 square and alternatingly adding and substrating squares of areas 1/3, 1/5, 1/7, etc. from the original square in a way that we keep approximating the quarter circle.
Still, any other intuitive proof on why the alternate sum of reciprocals of odd numbers is related to $\pi$ would be great to have.