# 任意两个自然数互质的概率？

$\sum_{d=1}^{\infty} \frac{P}{d^2} = 1$

$P = \left(\sum_{d=1}^{\infty} \frac{1}{d^2} \right)^{-1} = \frac{1}{\zeta(2)} = \left(\frac{\pi^2}{6}\right) ^{-1} = \frac{6}{\pi^2}.$

$\zeta(s) = \sum_{n=1}^\infty n^{-s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$

$\lim_{n\to\infty} P_k(n) = \frac{1}{\zeta(k)}$

$\prod_p \left(1-\frac{1}{p^2}\right) = \left( \prod_p \frac{1}{1-p^2} \right)^{-1} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}.$

[1] Coprime – Probabilities. (n.d.). In Wikipedia. Retrieved July 25, 2013, from https://en.wikipedia.org/wiki/Coprime#Probabilities.
[2] Hardy, G. Godfrey Harold, and Edward M. Wright. An introduction to the theory of numbers. Oxford University Press, 1979.