Coprimes 1

This is a classic number theory problem. Two numbers have no factor in common if and only if they have no prime factor in common. For two fixed integers, the probability that one of them has a prime $p$ as a factor is $\frac{1}{p}$ . Thus, the probability that they share $p$ as a factor is $\frac{1}{p^2}$ and the probability that they don't have common factor $p$ is $1-\frac{1}{p^2}$ . The probability that they have no prime factors in common is simply the product over all primes $p$ :

$\begin{aligned} \prod_{p\text{ prime}} 1-\frac{1}{p^2} &= \prod_{p\text{ prime}} \left(\frac{1}{1-\frac{1}{p^2}}\right)^{-1} \\ &= \left(\prod_{p\text{ prime}}\left(\sum_{n=1}^\infty \frac{1}{p^{2n}}\right)\right)^{-1} && \color{#3D99F6}\frac{1}{1-\frac{1}{p^2}} \text{ is the closed form of a geometric series}\\ &= \left( \prod_{p\text{ prime}} \left( 1 + \frac{1}{p^2} + \frac{1}{p^4} + \cdots \right)\right)^{-1} \\ &= \left( \left( 1 + \frac{1}{2^2} + \frac{1}{2^4} + \cdots \right) \left( 1 + \frac{1}{3^2} + \frac{1}{3^4} + \cdots \right) \left( 1 + \frac{1}{5^2} + \frac{1}{5^4} + \cdots \right) \right)^{-1} \\ &= \left(1+\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots \right)^{-1} && \color{#3D99F6}(1) \text{ By the Fundamental Theorem of Arithmetic} \\ &= \left(\frac{\pi^2}{6}\right)^{-1} && \color{#3D99F6}(2) \text{ By the Basel Problem} \\ &= \frac{6}{\pi^2} \\ &\approx \boxed{0.6079} \end{aligned}$

(1) When the expression in the preceding line is expanded, the product $\dfrac{1}{p_1^{2m_1}p_2^{2m_2}\cdots} = \dfrac{1}{(p_1^{m_1}p_2^{m_2}\cdots)^2}$ (where $p_k$ denotes the $k$ th prime) will appear precisely once for each $(m_1,m_2,\ldots)\in\mathbb{N}^\infty$ . By the Fundamental Theorem of Arithmetic, these terms correspond precisely to the natural numbers, and $\frac{1}{n^2}$ will appear in the sum precisely once for each $n\in\mathbb{N}$ .

(2) Finding a closed form for this sum is the famous Basel Problem , first solved by Leonard Euler in 1734. The most straightforward proof involves manipulating the Taylor Series for $\frac{\sin x}{x}$ , and can be found on the linked page.