# Does the sum $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converge?

Prove, if possible in an elementary way, that $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converges/diverges, where $p_n$ denotes the $n^{\textrm{th}}$ prime.

• Ummm... people don't like it when you use the imperative voice "Prove X". People do like it if you give context for the problem - how did it arise? etc. Feb 4 '16 at 4:06
• I essentially stumbled on this problem by accident. It seems interesting primarily as any "trivial" elementary bounds appear to fail, but using more machinery like Brun's estimates appears non-trivial. Feb 4 '16 at 4:10
• The converges/diverges refers to proving that it either converges OR it diverges. Still point well taken for the future. Feb 4 '16 at 4:11
• The convergence of your series follows from upper bound sieve estimates. In fact, a somewhat stronger result is shown in this paper: Erdös, Paul(H-AOS); Nathanson, Melvyn B.(1-CUNY7) On the sum of the reciprocals of the differences between consecutive primes. Number theory (New York, 1991–1995), 97–101, Springer, New York, 1996. Feb 4 '16 at 4:23
• I was about to say that the series "should" converge using crude probabilistic heuristics: If $p_n$ is a prime, the next integers should have something like a $1/\log p_n$ probability of being prime. So the expected value of $1/(p_{n+1}-p_n)$ given $p_n$ should be something like $\log\log p_n/\log p_n$. Now the sum is something like $\sum \frac{\log\log n}{n\log^2n}$. Of course now I should check how this checks out in the Erdös-Nathanson paper. Feb 4 '16 at 4:28

the authors show that $$\sum_{n=3}^{\infty} \frac{1}{n(\log\log{n})^c (p_{n+1}-p_n)}$$ converges for every choice of $c>2$. (I start the sum at $n=3$ rather than $n=2$ as as is done in the paper to avoid the annoyance that $\log\log{2} < 0$.) They also present a heuristic argument that the series diverges when $c=2$. The main tool in the proof is Brun's classical upper bound sieve estimate for the number of prime pairs $p, p+N$ with $p$ below a given bound.
Since $p_n$ is bounded below by a positive constant multiple of $n\log{n}$, the convergence of your series follows by comparison.