Euler’s Product Form of Riemann Zeta Function

Riemann zeta function is a rather simple-looking function. For any number s, the zeta function \zeta(s) is the sum of the reciprocals of all natural numbers raised to the s^\mathrm{th} power.

\begin{array}{ccl}\zeta(s)&=&\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n^s}}\\&=&1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\cdots\end{array}

Although the variable s is a complex number, for the sake of simplicity, we will treat s as real.  (Real numbers are a subset of complex numbers.)

Now let us multiply both sides of the above equation with \frac{1}{2^s}, which gives us

\begin{array}{ccl}\frac{1}{2^s}\zeta(s)&=&\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+\frac{1}{12^s}+\cdots\end{array}

Now let us subtract the second equation from the first one, getting

\begin{array}{ccl}(1-\frac{1}{2^s})\zeta(s)&=&1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\frac{1}{12^s}+\cdots\end{array}

Thus we have removed all  terms of the form \frac{1}{(2k)^s} from the right hand side, for k=1,2,3,\cdots. Next, if we multiply the above equation with \frac{1}{3^s} and then subtract the result from the first equation, all terms of the form \frac{1}{(3k)^s},k=1,2,3,\cdots will be from the right hand side, and we will get

\begin{array}{ccl}(1-\frac{1}{3^s})(1-\frac{1}{2^s})\zeta(s)&=&1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+\frac{1}{19^s}+\cdots\end{array}

We can continue in this fashion with all remaining prime numbers 5,7,11,13,\cdots. In the end, we will get

\begin{array}{ccl}\cdots (1-\frac{1}{13^s})(1-\frac{1}{11^s})(1-\frac{1}{7^s})(1-\frac{1}{5^s})(1-\frac{1}{3^s})(1-\frac{1}{2^s})\zeta(s)&=&1\end{array}

which implies

\begin{array}{rll}\displaystyle \prod_{p=2,3,5,7,11,13,\cdots}^{\infty}{\left(1-\frac{1}{p^s}\right)}\zeta(s)&=&1\\\implies \displaystyle\prod_{p \in \text{Prime}}{\left(1-\frac{1}{p^s}\right)}\zeta(s)&=&1\\\implies \zeta(s)&=&\frac{1}{\displaystyle\prod_{p \in \text{Prime}}{\left(1-\frac{1}{p^s}\right)}}\\\implies\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n^s}}&=&\displaystyle\prod_{p \in \text{Prime}}{(1-\frac{1}{p^s})^{-1}}\end{array}

Thus we come to this wonderful relationship. On one side, powers of all natural numbers, in sum form. On the other side, powers of all prime numbers, in product form. And they are equal! You can take reciprocals of all prime numbers, raise each to the power s, subtract each from 1, multiply them all … and you get the sum of the reciprocals of all natural numbers, each raised to the same power. Primes are the building blocks of natural numbers.

If this is not magic, magic does not exist.

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