# Divergence of sum reciprocal of primes using Bertrand's Postulate

I have been trying to prove that the series of reciprocal of primes diverges by only using Bertrand's Postulate. Does anybody know if this is possible? Or is it the case that this postulate is not strong enough to imply it?

Thanks!

• Could you compare with the infinite sum of 1/(2n)? – TomGrubb Apr 26 '15 at 22:15

Bertrand's postulate for a sequence $a_n$ (i.e. for every positive integer $k>1$ there is a member of the sequence with $k \le a_n < 2 k$) would be satisfied for e.g. the sequence $a_n = 2^n$, and the sum of its reciprocals converges.

Bertrand's postulate and Cauchy's condensation are not enough. The first one gives that between $N+1$ and $2N$ there is at least one prime, hence:

$$\sum_{\substack{p\text{ prime}\\p\leq 2^M}}\frac{1}{p}\geq \sum_{m=1}^{M-1}\frac{1}{2^m},\tag{1}$$ but the last series is converging. This is not surprising, since also the powers of two have the "Bertrand postulate property" but the sum of their reciprocals is bounded.

• No. Bertrand's postulate says that there is at least one prime between n and 2n. See Robert Israel's answer for the correct usage. – marty cohen Apr 26 '15 at 22:58
• @martycohen: yes, sure, now fixed. – Jack D'Aurizio Apr 26 '15 at 23:00

It appears the first valid proof of divergence did happen before the prime number theorem by about 22 years, http://en.wikipedia.org/wiki/Mertens%27_theorems#Mertens.27_second_theorem_and_the_prime_number_theorem

Let's see, very long before Hoheisel's result, which is a bound (possibly the first such) on prime gaps that immediately shows the sum of prime reciprocals diverges. http://en.wikipedia.org/wiki/Prime_gap#Upper_bounds

I'm trying to decide how PNT shows divergence... seems very likely