Asymptotic behavior of $\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-p_k}$

Question

I refer to my previous question Asymptotic behavior of a certain sum of ratios of consecutives primes. We can split the sum $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}$$ where $p_k$ stands for the prime of index $k$, into the following two

$\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-\,p_k}$ ~ $\frac{n\,(n+1)}{e}\,\log\log n$

$\sum_{k=1}^{n}\frac{p_{k}}{p_{k+1}-\,p_k}$ ~ $\frac{(n-1)\,n}{e}\,\log\log n$

Is there anybody who can confirm this asymptotic behavior and, if it is correct, give a sketch of a proof?

Answer 1

My response to your earlier question applies almost verbatim. The heuristic reasoning there gives that \begin{align*} \sum_{k=1}^{n}\frac{p_k}{p_{k+1}-p_k}&\sim\frac{C}{2}\, n^2\log\log n,\\ \sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-p_k}&\sim\frac{C}{2}\, n^2\log\log n, \end{align*} where the constant $C>0$ is the same as in that post. As I wrote there, this constant is almost surely different from $2/e$. In fact, as Lucia kindly pointed out in a comment, $C=1$.

The difficulty in estimating these sums lies in the erratic behaviour of the denominator $p_{k+1}-p_k$. The numerator $p_k$ (resp. $p_{k+1}$ or $p_k+p_{k+1}$) is easy to handle as it is asymptotically $k\log k$ (resp. $k\log k$ or $2k\log k$).