Dear Uncle Colin,

Why does $1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4}$?

My Arithmetic Doesn’t Have A Viable Answer

There are several ways to show this, but I think my favourite stems from the Maclaurin series for $\ln(1+x)$. Or rather, in this case, $\ln(1+ix)$:

$\ln(1+ix) = ix + \frac{1}{2}x^2 - i\frac{1}{3}x^3 - \frac{1}{4}x^4 - \dots$, for $|x|<1$.

As $x$ approaches 1, this gives $\ln(1+x) \to \br{\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\dots} + i\br{1 - \frac{1}{3}+\frac{1}{5}-\dots}$.

The real part is (in the limit) $\frac{1}{2}\br{1 -\frac{1}{2}+\frac{1}{3}-\dots}$, which works out to br $\frac{1}{2}\ln(2)$ (again, in the limit).

The imaginary part is our left hand side.

However, we know what $\ln(1+i)$ is – we can write $1+i$ in polar form as $\sqrt{2}e^{i\pi/4}$, so $\ln(1+i) = \frac{1}{2}\ln(2) + i \frac{\pi}{4}$ ((At least, on the principal branch.))

The imaginary part of that is equal to the imaginary part of the expansion, so $1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{\pi}{4}$.

(What do you mean, handwavey?) Hope that helps!

- Uncle Colin