Coprimes 2

For more detail, see the solution to the two integer case, as the proof is virtually identical.

---

Three integers have no common factor if an only if they have no common prime factor. For three fixed integers and a given prime $p$ , this has probability $1-\frac{1}{p^3}$ . Thus, we compute

$\begin{aligned} \prod_{p\text{ prime}} 1-\frac{1}{p^3} &= \prod_{p\text{ prime}} \left(\frac{1}{1-\frac{1}{p^3}}\right)^{-1} \\ &= \left(\prod_{p\text{ prime}}\left(\sum_{n=1}^\infty \frac{1}{p^{3n}}\right)\right)^{-1}\\ &= \left( \prod_{p\text{ prime}} \left( 1 + \frac{1}{p^3} + \frac{1}{p^6} + \cdots \right)\right)^{-1} \\ &= \left( \left( 1 + \frac{1}{2^3} + \frac{1}{2^6} + \cdots \right) \left( 1 + \frac{1}{3^3} + \frac{1}{3^6} + \cdots \right) \left( 1 + \frac{1}{5^3} + \frac{1}{5^6} + \cdots \right) \right)^{-1} \\ &= \left(1+\frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \cdots \right)^{-1}\\ &= \left(\sum_{n=1}^\infty\frac{1}{n^3}\right)^{-1} \\ &= \left(\zeta(3)\right)^{-1} &&\color{#3D99F6}\text{Where }\zeta(s)\text{ is the Reimann zeta function}\\ &\approx (1.202)^{-1} && \color{#3D99F6}(1)\text{ Apéry's constant} \\ &\approx \boxed{0.8319} \end{aligned}$

(1) To my knowledge, there is no straightforward way to compute $\zeta(3)$ , otherwise known as Apéry's constant.