# From a Sum to a Product

In The Basel Problem we saw that Euler answered the question by showing that the sum of the reciprocals of the square numbers is equal to $$\pi^2/6$$.

That is

$$\displaystyle \sum_{n=1}^\infty \frac {1}{n^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + … = \frac{\pi^2}{6}$$

The partial sums slowly converge to the limit of $$1.644 934 066 …$$.

It was a remarkable proof that utilised the Taylor/Maclaurin series of the sine function, hence the appearance of $$\pi$$.

$$\displaystyle \sin(x) \approx x-\frac{x^3}{3!} + \frac{x^5}{5!}-\frac{x^7}{7!} + …$$

Two years later, in $$1737$$, Euler published a paper proving that this infinite sum over all the natural numbers, $$\Bbb N$$, is equal to an infinite product over all the prime numbers, $$\Bbb P$$.

This is the first key to the riddle of the primes.

Once again Euler shows his dexterity, this time by using the sieve of Eratosthenes.

We have :

$$\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + …$$

Multiply both sides by $$1/2^2$$ :

$$\displaystyle \frac{\pi^2}{6}\cdot\frac{1}{2^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \frac{1}{10^2} + \frac{1}{12^2} + \frac{1}{14^2} + …$$

Subtract the second equation from the first equation:

LHS :

$$\displaystyle \frac{\pi^2}{6}-\frac{\pi^2}{6}\cdot\frac{1}{2^2} = \frac{\pi^2}{6}\left(1-\frac{1}{2^2}\right)$$

RHS :

$$\displaystyle \left(1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + …\right)-\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} + \frac{1}{10^2} + … \right)$$

$$\displaystyle = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \frac{1}{13^2} + …$$

Thus :

$$\displaystyle \frac{\pi^2}{6}\left(1-\frac{1}{2^2}\right) = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \frac{1}{13^2} + …$$

We repeat these two steps for all the prime numbers :

$$\displaystyle \frac{\pi^2}{6}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right) … = 1$$

The RHS has been sieved; the primes are on the LHS and the multiples of primes have disappeared.

Solving for $$\pi^2/6$$ gives :

$$\displaystyle \frac{\pi^2}{6} = \frac{1}{\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right) …}$$

This is the Euler product (there are others).

We write it as :

$$\bbox[15px,border:2px solid black] { \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} = \prod_p \frac{1}{\left(1-\frac{1}{p^2}\right)} }$$

The Basel Problem

© OldTrout $$2020$$