# How would you prove that this sum involving prime gaps has a limit?

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?

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 $$\Re\phi(n)\sim n-1.$$