I am looking for the asymptotic growth of the following 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 $k$.


Manual computations show, for small values of n, a behavior quite similar to that of the sum over naturals
$$\sum\_{k=0}^{n-1}(2k+1)=n^2$$
But more accurate simulations with Python suggest that


$\sum\_{k=1}^{n}\frac{p\_{k+1}+\,p\_k}{p\_{k+1}-\,p\_k}$ ~ $\frac{2}{e}\,n^2\log\log n$


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



Asymptotic behavior of a certain sum of ratios of consecutives primes
It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p\_{k+1}-p\_k$, on which we only have conjectures (see also my Added section below). 


It is clear that
$$\#\{k\leq n: p\_{k+1}-p\_k>\log n\}<\frac{p\_{n+1}}{\log n},$$
hence the contribution of $p\_{k+1}-p\_k>\log n$ is 
$$\sum\_{\substack{1\leq k\leq n\\p\_{k+1}-p\_k>\log n}}\frac{p\_{k+1}+p\_k}{p\_{k+1}-p\_k}<\frac{p\_{n+1}}{\log n}\cdot\frac{p\_{n+1}+p\_n}{\log n}=O(n^2).$$
Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that
$$\#\{k\leq n: p\_{k+1}-p\_k=h\}\sim\frac{n}{\log n}\cdot 2C\_2\cdot D\_h,\tag{$\ast$}$$
where
$$C\_2:=\prod\_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D\_h:=\prod\_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$
If we believe in this, then integration by parts gives that the contribution of $p\_{k+1}-p\_k=h$ is asymptotically $n^2 C\_2 D\_h/h$. Based on this heuristic, it is reasonable to conjecture that
$$\sum\_{k=1}^{n}\frac{p\_{k+1}+p\_k}{p\_{k+1}-p\_k}\sim C\_2\, n^2\sum\_{h\leq\log n}\frac{D\_h}{h}.$$
It is straightforward that the Dirichlet series of $D\_h$ factors as
$$\sum\_{h=1}^\infty\frac{D\_h}{h^s}=\zeta(s)F(s),$$
where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically,
$$\sum\_{k=1}^{n}\frac{p\_{k+1}+p\_k}{p\_{k+1}-p\_k}\sim C\, n^2\log\log n,$$
where $C:=C\_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this.


**Added.** It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$. As Lucia kindly pointed out, a result of Gallagher's implies that $C=1$.


