Start at Basel Problem

In 18th century, Euler solved a well known problem, even Leibniz cannot solve it, either. This event made him become famous and confident. Basel problem shows \(\displaystyle \lim_{n \to +\infty} \left(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{n^2}\right)= \sum_{m=1}^{\infty} \dfrac{1}{m^2}=\dfrac{\pi^2}{6}\). It is a wonderful theory, but I’m not satisfied with this. We can use it to find more interesting conclusions. \[\begin{align*} \because \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2} \times \dfrac{1}{1^2}+\dfrac{1}{2^2} \times \dfrac{1}{2^2}+\dfrac{1}{2^2} \times \dfrac{1}{3^2}+\cdots \\ \therefore \dfrac{1}{2^2} \times \dfrac{\pi^2}{6} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore \left(1-\dfrac{1}{2^2}\right) \dfrac{\pi^2}{6} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \therefore \dfrac{\pi^2}{8} &=\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots \\ \dfrac{\pi^2}{24} &=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\cdots \\ \therefore \left(\dfrac{1}{8}-\dfrac{1}{24}\right)\pi^2 &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \therefore \dfrac{\pi^2}{12} &=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\cdots \\ \end{align*}\] It’s true that Basel problem has an important relationship with Riemann \(\zeta\) function, because Basel problem is in order to find \(\zeta(2)\). About 100 years later, Riemann used one equation to build a bridge between functions and prime numbers. The equation is \(\displaystyle \zeta(s) =\sum_{n=1} ^{\infty} \dfrac{1}{n^s}= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1}\). Here is Euler's proof. \[\begin{align*} \because \zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{2^s}+\dfrac{1}{3^s}+\cdots \\ \therefore \dfrac{1}{2^s} \zeta(s) &=\dfrac{1}{2^s}+\dfrac{1}{4^s}+\dfrac{1}{6^s}+\cdots \\ \therefore \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{3^s}+\dfrac{1}{5^s}+\cdots \\ \because \dfrac{1}{3^s} \times \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{3^s} \times \dfrac{1}{1^s}+\dfrac{1}{3^s} \times \dfrac{1}{3^s}+\dfrac{1}{3^s} \times \dfrac{1}{5^s}+\cdots \\ \therefore \left(1- \dfrac{1}{3^s}\right) \left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{5^s}+\dfrac{1}{7^s}+\cdots \\ \therefore \left(1-\dfrac{1}{5^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{2^s}\right)\zeta(s) &=\dfrac{1}{1^s}+\dfrac{1}{7^s}+\dfrac{1}{11^s}+\cdots \\ \therefore 1 &=\zeta(s)\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\cdots \\ \therefore \zeta(s) &=\dfrac{1}{\left(1-\dfrac{1}{2^s}\right)\left(1- \dfrac{1}{3^s}\right)\left(1-\dfrac{1}{5^s}\right)\left(1-\dfrac{1}{7^s}\right)\cdots} \\ \therefore \sum_{n=1} ^{\infty} \dfrac{1}{n^s} &= \prod_{p \text{ prime}} ^{\infty} (1-p^{-s})^{-1} \end{align*}\]