Definitions

# Proof of the Euler product formula for the Riemann zeta function

We will prove that the following formula holds:

begin{align} zeta(s) & = 1+frac{1}{2^s}+frac{1}{3^s}+frac{1}{4^s}+frac{1}{5^s}+ cdots & = prod_{p} frac{1}{1-p^{-s}} end{align}

where ζ denotes the Riemann zeta function and the product extends over all prime numbers p.

### Proof of the Euler product formula

This sketch of a proof only makes use of simple algebra that most high school students can understand. This was originally the method by which Euler discovered the formula. There is a certain sieving property that we can use to our advantage:

$zeta\left(s\right) = 1+frac\left\{1\right\}\left\{2^s\right\}+frac\left\{1\right\}\left\{3^s\right\}+frac\left\{1\right\}\left\{4^s\right\}+frac\left\{1\right\}\left\{5^s\right\}+ cdots$

$frac\left\{1\right\}\left\{2^s\right\}zeta\left(s\right) = frac\left\{1\right\}\left\{2^s\right\}+frac\left\{1\right\}\left\{4^s\right\}+frac\left\{1\right\}\left\{6^s\right\}+frac\left\{1\right\}\left\{8^s\right\}+frac\left\{1\right\}\left\{10^s\right\}+ cdots$

Subtracting the second from the first we remove all elements that have a factor of 2:

$left\left(1-frac\left\{1\right\}\left\{2^s\right\}right\right)zeta\left(s\right) = 1+frac\left\{1\right\}\left\{3^s\right\}+frac\left\{1\right\}\left\{5^s\right\}+frac\left\{1\right\}\left\{7^s\right\}+frac\left\{1\right\}\left\{9^s\right\}+frac\left\{1\right\}\left\{11^s\right\}+frac\left\{1\right\}\left\{13^s\right\}+ cdots$

Repeating for the next term:

$frac\left\{1\right\}\left\{3^s\right\}left\left(1-frac\left\{1\right\}\left\{2^s\right\}right\right)zeta\left(s\right) = frac\left\{1\right\}\left\{3^s\right\}+frac\left\{1\right\}\left\{9^s\right\}+frac\left\{1\right\}\left\{15^s\right\}+frac\left\{1\right\}\left\{21^s\right\}+frac\left\{1\right\}\left\{27^s\right\}+frac\left\{1\right\}\left\{33^s\right\}+ cdots$

Subtracting again we get:

$left\left(1-frac\left\{1\right\}\left\{3^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{2^s\right\}right\right)zeta\left(s\right) = 1+frac\left\{1\right\}\left\{5^s\right\}+frac\left\{1\right\}\left\{7^s\right\}+frac\left\{1\right\}\left\{11^s\right\}+frac\left\{1\right\}\left\{13^s\right\}+frac\left\{1\right\}\left\{17^s\right\}+ cdots$

where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely we get:

$cdots left\left(1-frac\left\{1\right\}\left\{11^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{7^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{5^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{3^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{2^s\right\}right\right)zeta\left(s\right) = 1$

Dividing both sides by everything but the $zeta\left(s\right)$ we obtain:

$zeta\left(s\right) = frac\left\{1\right\}\left\{left\left(1-frac\left\{1\right\}\left\{2^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{3^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{5^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{7^s\right\}right\right)left\left(1-frac\left\{1\right\}\left\{11^s\right\}right\right) cdots \right\}$

This can be written more concisely as an infinite product over all primes p:

$zeta\left(s\right);=;prod_\left\{p\right\} \left(1-p^\left\{-s\right\}\right)^\left\{-1\right\}.$

To make this proof rigorous, we need only observe that when Re(s) > 1, the sieved right-hand side approaches 1, which follows immediately from the convergence of the Dirichlet series for $zeta\left(s\right)$.

An interesting result can be found for $zeta\left(1\right)$

$cdots left\left(1-frac\left\{1\right\}\left\{11\right\}right\right)left\left(1-frac\left\{1\right\}\left\{7\right\}right\right)left\left(1-frac\left\{1\right\}\left\{5\right\}right\right)left\left(1-frac\left\{1\right\}\left\{3\right\}right\right)left\left(1-frac\left\{1\right\}\left\{2\right\}right\right)zeta\left(1\right) = 1$
which can also be written as,
$cdots left\left(frac\left\{10\right\}\left\{11\right\}right\right)left\left(frac\left\{6\right\}\left\{7\right\}right\right)left\left(frac\left\{4\right\}\left\{5\right\}right\right)left\left(frac\left\{2\right\}\left\{3\right\}right\right)left\left(frac\left\{1\right\}\left\{2\right\}right\right)zeta\left(1\right) = 1$
which is,
$left\left(frac\left\{... 10.6.4.2.1\right\}\left\{... 11.7.5.3.2\right\}right\right)zeta\left(1\right) = 1$
as, $zeta\left(1\right) = 1+frac\left\{1\right\}\left\{2\right\}+frac\left\{1\right\}\left\{3\right\}+frac\left\{1\right\}\left\{4\right\}+frac\left\{1\right\}\left\{5\right\}+ cdots$

thus,

$1+frac\left\{1\right\}\left\{2\right\}+frac\left\{1\right\}\left\{3\right\}+frac\left\{1\right\}\left\{4\right\}+frac\left\{1\right\}\left\{5\right\}+ cdots ;=; left\left(frac\left\{... 11.7.5.3.2\right\}\left\{... 10.6.4.2.1\right\}right\right)$

We know that the left-hand side of the equation diverges to infinity therefore the numerator on the right-hand side (the series of primes) must also be infinite for divergence.

### Another proof

Each factor (for a given prime p) in the product above can be expanded to a geometric series consisting of the reciprocal of p raised to multiples of s, as follows

$frac\left\{1\right\}\left\{1-p^\left\{-s\right\}\right\} = 1 + frac\left\{1\right\}\left\{p^s\right\} + frac\left\{1\right\}\left\{p^\left\{2s\right\}\right\} + frac\left\{1\right\}\left\{p^\left\{3s\right\}\right\} + cdots + frac\left\{1\right\}\left\{p^\left\{ks\right\}\right\} + cdots$

When $zeta\left(s\right) > 1$, $left|p^\left\{-s\right\}right| < 1$ and this series converges absolutely. Hence we may take a finite number of factors, multiply them together, and rearrange terms. Taking all the primes p up to some prime number limit q, we have

$left|zeta\left(s\right) - prod_\left\{p le q\right\}left\left(frac\left\{1\right\}\left\{1-p^\left\{-s\right\}\right\}right\right)right| < sum_\left\{n=q+1\right\}^infty frac\left\{1\right\}\left\{n^sigma\right\}$

where σ is the real part of s. By the fundamental theorem of arithmetic, the partial product when expanded out gives a sum consisting of those terms $frac\left\{1\right\}\left\{n^s\right\}$ where n is a product of primes less than or equal to q. The inequality results from the fact that therefore only integers larger than q can fail to appear in this expanded out partial product. Since the difference between the partial product and ζ(s) goes to zero when σ > 1, we have convergence in this region.

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