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.



Does the sum $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converge?
By popular demand, I am converting my comment above to an answer. Yes, the series converges. In the 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


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.


