Calling $P$ the set of prime numbers, Euler proved that $$\sum_{n=1}^{\infty}\frac 1{n^s}=\zeta (s)=\prod_{p\in P} \frac 1{1-\frac 1{p^s}}$$ At University my professor proved that in this way:
- First we can see $$\frac 1{1-\frac 1{p^s}}=\sum_{k=0}^{\infty} \left(\frac 1{p^s}\right)^k=\sum_{k=0}^{\infty}\frac 1{p^{ks}}$$
- Then $$\prod_{p\in P} \frac 1{1-\frac 1{p^s}}= \prod_{p\in P} \sum_{k=0}^{\infty}\frac 1{p^{ks}}$$
- So we see that when we calculate the product of the sums we obtain a sum of reciprocal of $s$-th powers of natural numbers, and for all $s$-th power of a natural number then it’s reciprocal is in the sum... so we conclude thesis.
This demonstration is surely clear but it appears to me a bit “informal” at the moment of the conclusions... Does anyone know a different demonstration of this Identity?
