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

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?

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 $$2k\log k$$).