I refer to my previous question [Asymptotic behavior of a certain sum of ratios of consecutives primes](https://mathoverflow.net/questions/351875/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?



Asymptotic behavior of $\sum_{k=1}^{n}\frac{p_{k+1}}{p_{k+1}-p_k}$
My [response](https://mathoverflow.net/questions/351875/asymptotic-behavior-of-a-certain-sum-of-ratios-of-consecutives-primes) 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 $2k\log k$).


