-3
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $2k\log k$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.