<s>[INST] Geometric intuition for $\pi /4 = 1 - 1/3 + 1/5 - \cdots$? Following reading this great post : Interesting and unexpected applications of $\pi$, Vadim's answer reminded me of something an analysis professor had told me when I was an undergrad - that no one had ever given him a satisfactory intuitive explaination for why this series definition should be true (the implied assumption being that it is so simple, there should be some way to look at this to make it intuitive).
Now it follows simply from putting $x=1$ in the series expansion $$ \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots $$
but I don't see anything geometrically intuitive about this formula either!
 [/INST] By elementary means:
In the figure below, we have drawn a quarter circle inside a unit square and divided the top side in $n$ equal parts. Then we have chosen the height of the blue rectangles in such a way that the area of a rectangle equals the area of the corresponding sector (the aperture angles go decreasing).

If you look at the enhanced triangle, its long sides have length $\sqrt{1+x^2}$, by Pythagoras, so that its short side is $\sqrt{1+x^2}$ times longer than the arc on the circle ($x$ horizontal in $[0,1]$). Then by similarity of triangles, the ratio of the width of a rectangle to this short side is also $\sqrt{1+x^2}$, and we conclude that the equation of the green curve is 
$$y(x)=\frac1{2(1+x^2)}.$$
Furthermore, the area under the curve is the same as the area of the eighth of a circle, i.e. $\dfrac\pi8$.
Now the (doubled) area is obtained as the sum of the areas of the rectangles,
$$\frac\pi4=2\sum_{k=1}^n\frac1ny\left(\frac kn\right)=\frac1n\sum_{k=1}^n\frac1{1+\dfrac{k^2}{n^2}}.$$
To evaluate it $^{(1)}$, we use the identity
$$\frac1{1+\dfrac{k^2}{n^2}}=1-\frac{k^2}{n^2}+\frac{k^4}{n^4}-\frac{k^6}{n^5}+\cdots$$
and by the Faulhaber formula $^{(2)}$, we know that the sum of the $n$ first $p^{th}$ powers is $\dfrac{n^{p+1}}{p+1}$.
This gives
$$\frac\pi4=\frac1n\left(n-\frac{n^3}{3n^2}+\frac{n^5}{5n^4}-\frac{n^7}{7n^6}+\cdots\right)$$ which is the claimed formula.

$(1)$ This is "justified" by
$$(1+x)(1-x+x^2-x^3+x^4-\cdots)=1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5\cdots=1\pm x^\infty$$ where the last term vanishes when $x<1$.

$(2)$ This is "justified", using the binomial theorem, by
$$n^p=\sum_{k=1}^n k^p-\sum_{k=1}^{n-1} k^p\approx\frac{n^{p+1}}{p+1}-\frac{(n-1)^{p+1}}{p+1}=\frac{n^{p+1}-(n^{p+1}-(p+1)n^p+\binom{p+1}2n^{p-1}-\binom{p+1}3n^{p-2}+\cdots)}{p+1}=n^p+\cdots.$$
The neglected terms are of lower degree in $n$ and can be ignored for large $n$.
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