∑n 1/ns = ∏p ps/(ps − 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/ps).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/q2 + ... So we have ps/(ps − 1) = ∑n 1/pns, where the sum is over the nonnegative integers. Expanding the right side of the identity, we have ∑n 1/pns = 1 + 1/ps + 1/p2s + 1/p3s + ... It follows that ∏p ∑n 1/pns = (1 + 1/2s + 1/22s + 1/23s + ...)(1 + 1/3s + 1/32s + 1/33s + ...)(1 + 1/5s + 1/52s + 1/53s + ...) ... Such an infinite product is unwieldy, to be sure, but we observe that a general term in its expansion is of the form 1/2as·3bs·5cs·... = 1/(2a·3b·5c·...)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 2a·3b·5c·... So we have ∑n 1/ns = ∏p ps/(ps − 1).