For a real number s, a positive integer n, and a prime number p, we have the astonishing result that

∑_{n} 1/n^{s} = ∏_{p} p^{s}/(p^{s} − 1)

That is, a sum over the integers is equal to a product over the primes. This result, attributed to the prolific Swiss mathematician Leonhard Euler [1707-1783], is crucial to much of the analysis and number theory that has followed.

Note: The product is sometimes written as ∏_{p} 1/(1 − 1/p^{s}).

Proof: Observe that both the product and the sum converge for all s > 1.

Next, recall the identity q/(q − 1) = 1 + 1/q + 1/q^{2} + ...

So we have p^{s}/(p^{s} − 1) = ∑_{n} 1/p^{ns}, where the sum is over the nonnegative integers.

Expanding the right side of the identity, we have

∑_{n} 1/p^{ns} = 1 + 1/p^{s} + 1/p^{2s} + 1/p^{3s} + ...

It follows that

∏_{p} ∑_{n} 1/p^{ns} = (1 + 1/2^{s} + 1/2^{2s} + 1/2^{3s} + ...)(1 + 1/3^{s} + 1/3^{2s} + 1/3^{3s} + ...)(1 + 1/5^{s} + 1/5^{2s} + 1/5^{3s} + ...) ...

Such an infinite product is unwieldy, to be sure, but we observe that a general term in its expansion is of the form 1/2^{as}·3^{bs}·5^{cs}·... = 1/(2^{a}·3^{b}·5^{c}·...)^{s}, where a, b, c, ... are non-negative integers.

We now use the Fundamental Theorem of Arithmetic, which says that we can uniquely express each integer as a product of primes. That is, there is a one-to-one correspondence between the positive integers and the set of all products of the form 2^{a}·3^{b}·5^{c}·...

So we have ∑_{n} 1/n^{s} = ∏_{p} p^{s}/(p^{s} − 1).

QED