The Riemann zeta function \(\zeta\) is a function of one complex variable \(s\) defined by \[\zeta(s)=\sum_{n=1}^{\infty} n^{-s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots.\] Note that \(\zeta(1)=\sum_{n=1}^{\infty} \frac{1}{n}=\infty\) and the Dirichlet series \(\sum_{n=1}^{\infty} n^{-s}\) converges if and only if \(\operatorname{Re}(s)>1\). Sometimes \(\zeta\) is defined as the analytic continuation of itself from the domain \(\{s\in \mathbb C\;|\;\operatorname{Re}(s)>1\}\) to \(\{s\in \mathbb C\;|\;s\neq 1\}\). The Riemann zeta function is an important topic of analytic number theory.

Theorem.(Euler product formula) \[\zeta(s)=\sum_{n=1}^{\infty} n^{-s}=\prod_{p\in \mathbb P}(1-p^{-s})^{-1}.\]

For an informal justification we look at Euler's original technique. First note that \[\frac{1}{2^s}\zeta(s)=\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\cdots.\] Subtracting it from \(\zeta(s)\), we get \[\left(1-\frac{1}{2^s}\right)\zeta(s)=1+\frac{1}{3^s}+\frac{1}{5^s}+\cdots,\] which has no term of the form \(\frac{1}{(2k)^{s}}\), \(k=1,2,\ldots\). Similarly we get \[\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s)=1+\frac{1}{5^s}+\frac{1}{7^s}+\cdots,\] which has no term of the form \(\frac{1}{(2k)^{s}}\) and \(\frac{1}{(3k)^{s}}\), \(k=1,2,\ldots\). Continuing this process on primes, we get \[ \begin{align*} &\left(\prod_{p\in \mathbb P}(1-p^{-s})\right)\zeta(s)=\cdots\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s)=1\\ \implies &\zeta(s)=\left(\prod_{p\in \mathbb P}(1-p^{-s})\right)^{-1}=\prod_{p\in \mathbb P}(1-p^{-s})^{-1}. \end{align*} \]

Riemann Hypothesis (Bernhard Riemann, 1859): If \(s\) is a nontrivial root (i.e., a root that is not a negative even integer) of the analytical continuation of the Riemann zeta function, then \(s=\frac{1}{2}+ it\) for some real \(t\).

A valid proof or a counterexample of the Riemann hypothesis would get you a million dollars given by the Clay Mathematics Institute.

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