# Eikonal Blog

## 2016.08.24

### pi

Filed under: Uncategorized — sandokan65 @ 21:19

Seen somewhere recently:

• $\frac\pi4 = \frac11 - \frac13 + \frac15 - \frac17 + \cdots$
• $\frac{\pi^2}6 = \frac1{1^1} + \frac1{2^2} + \frac1{3^2} + \cdots$
• $\frac2\pi = \left(1-\frac13\right)\left(1+\frac15\right)\left(1-\frac17\right)\left(1-\frac1{11}\right)\left(1+\frac1{13}\right)\left(1+\frac1{17}\right)\cdots$ where fractional terms $\frac1p$ are over all primes, and the sign in front of them is minus if $p=4n-1$ and plus otherwise.