# Geometric intuition for π/4=1−1/3+1/5−⋯\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

I doubt that there is a simple geometric proof that can be visualized directly, like some circle area computation using some inclusion-exclusion procedure.

We could split the proof in three steps:

1. The area of a quarter of unit circle equals the area below $f(x)=\frac{1}{1+x^2}$ in the interval $[0,1]$

2. $\frac{1}{1+x^2}=1 - {x^2} + {x^4} - \cdots$ for $|x|<1$

3. $\int_0^1 x^{2n} dx= \frac{1}{2n+1}$

Step 1 admits several visual-geometric proofs (below I insert a diagram I made for the linked question, that shows the area equivalence for the full graph: $\int_{-\infty}^{\infty}1/(1+x^2) dx =\pi$; just trim it to the $[0,1]$ range).

But steps 2 and 3, though elementary from a analysis point of view, do not seem easy to visualize.

Explanation added by comment request: the graph (edited) intends to show that the two areas in yellow are (asymptotically) equal.

First, we have a circular sector with radius 1 and arc $s$, which is approximated by a triangle with the same height and base ; hence its area is $s/2$.

The other area is below the function $\frac{1}{2(1+x^2)}$, so its area is $\overline{CE} \times y = a \frac{1}{2(1+x^2)}$. We then compute $a$:

Because $B0C$ is rectangular $\ell=\overline{BC}=\sqrt{1+\overline{0C}^2}=\sqrt{1+x^2}$

Because triangles $CDB$ and $C'D'B$ are similar, $\ell=\frac{\overline{BC}}{\overline{B C^{'}}}=\frac{\overline{CD}}{\overline{C^{'}D^{'}}}=\frac{t}{s}$

Because $CED$ and $BOC$ are similar $\frac{a}{t}=\frac{\overline{CE}}{\overline{CD}}=\frac{\overline{BC}}{\overline{0B}}=\ell$

Hence $a= s \ell^2 = s (1+x^2)$ and the area is also $s/2$.

Applying this to the range $[0,1]$ we deduce that the area of an octave of a unit circle equals $\int_0^1 \frac{1}{2(1+x^2)}dx$