### How the Riemann zeta function is connected to prime numbers The Riemann zeta function is defined as $$ \zeta(s) = \sum_{n=1}^{+\infty}\frac{1}{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ We can multiply each side by $(1-2^{-s})$ $$ (1-2^{-s})\zeta(s) = \left(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots\right) (1-2^{-s}) $$ and distribute $$ (1-2^{-s})\zeta(s) = \left(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots\right) - \left(\frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{6^s} + \cdots\right) $$ to get $$ (1-2^{-s})\zeta(s) = \left(\frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{5^s} + \cdots\right) $$ Continuing with this thought, we can multiply each side by $(1-3^{-s})$ to get rid of more terms $$ (1-3^{-s})(1-2^{-s})\zeta(s) = \left(\frac{1}{1^s} + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{11^s} + \cdots\right) $$ We are just multiplying by $(1-p^{-s})$ where $p$ is prime, removing all terms whose denominators are divisible by $p^{s}$. If we keep going, there will be only one term left. We can denote this as $$ \prod_{p \text{ prime}}(1-p^{-s})\zeta(s) = \frac{1}{1^s} = 1 $$ or more simply $$ \zeta(s) = \prod_{p \text{ prime}}\frac{1}{1-p^{-s}} = \sum_{n=1}^{+\infty}\frac{1}{n^s} $$