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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}