# Infinite sum over the primes

I was recently trying to create my own unique proof that the harmonic series diverges, so what I did was realize that within the harmonic series is the reciprocal of powers of primes and for every prime, every reciprocal of the powers of that prime also are contained. By regrouping, you can show that for every prime, the powers of that prime add up to 1/(p-1) so we know that for every prime, the harmonic series at least contains 1/(p-1) which means that the harmonic series is greater than the sum from 1 to infinity of 1/(p-1) where the sum is over the primes. How would I go about showing that this sum diverges?

Sorry about formatting I am currently doing this on a mobile device.

• I appreciate your observation, but the sub-problem you have in mind seems a lot harder than the original problem itself, for the simple reason that it involves the distribution of the primes among natural numbers, which is something more difficult to grasp than some of the easier proofs that the Harmonic series diverges. Suppose you know that the primes are distributed asymptotically with density $\frac{x}{\log x}$ (and it's not easy to show), then you may be able to infer the convergence of the above sum using some integral inequality. – Teresa Lisbon Apr 20 '17 at 1:15
• You may be interested in reading about elementary proofs that the sum of the reciprocals of the primes diverges (that's more or less equivalent to what you want). Of course "elementary" is by no means a synonym for "easy". here is an introduction. – lulu Apr 20 '17 at 1:38