Let $p_k$ = the $k$th prime.

$$ \varphi(n) = \sum_{k=1}^{n-1} e^{i 2 \pi \frac{p_{k+1} - p_k}{p_n}} $$

seems to approach a constant point in $\Bbb{C}$ as $n \to \infty$. How can I prove it though?


I doubt the convergence.

By Bertrand's postulate, we will (except for small $n$) always have $p_{k+1}-p_k<\frac12 p_n$, which already gives all summands positive imaginary part. By known generalization of Bertrand, for each $\epsilon>0$, we have $p_{n+1}<(1+\epsilon)p_n$ for almost all $n$. Thus for $n$ large enough, all $\frac{p_{k+1}-p_k}{p_n}$ are small (between $0$ and $\frac16$, say), either beacuse $p_{k+1}$ is not much larger than $p_k$, or because $p_n$ is much larger than $p_{k+1}$. Then we have $n-1$ summands with each having real part $>\frac12$, say.

More precisely, the above shows that $$\Re\phi(n)\sim n-1. $$


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.