On the Sum of the Reciprocals of the Differences Between Consecutive Primes

The infinite series $$\sum\limits_{{n = 2}}^{\infty } {\frac{1}{{n{{{(\log \log n)}}^{c}}\log n}}}$$ converges if and only if c > 1. Let p_n denote the n-th prime number. By the prime number theorem, $$\sum\limits_{{i = 1}}^{n} {({{p}_{{i + 1}}} - {{p}_{i}}) = {{p}_{{n + 1}}} - 2 \sim n \log n,}$$ and so the difference between consecutive primes is on average logn. This suggests the question: For what values of c does the series $$\sum\limits_{{n = 2}}^{\infty } {\frac{1}{{n{{{(\log \log n)}}^{c}}({{p}_{{n + 1}}} - {{p}_{n}})}}}$$ converge? We shall prove convergence for c > 2, and give a heuristic argument why the series must diverge for c = 2.