Prove or refute that $\{p^{1/p}\}_{p\text{ prime}}$ to be equidistributed in $\mathbb{R}/\mathbb{Z}$ I've tried follow the Example 3 (see minute 30'40" of the reference), where is required the related Theorem (stated at minute 21') combined with Serre's formalism for $\mathbb{R}/\mathbb{Z}$ (also explained in the video) from a  video in YouTube, from the official channel matsciencechannel, An Introduction to Analytic Number Theory Lecture 07 Equidistribution by Ram Murty to ask myself 

Question. Is the sequence $\{p^{1/p}\}$, as $p$ varies over all primes, equidistributed in $\mathbb{R}/\mathbb{Z}$?

I need to check if $$\frac{1}{\pi(N)}\sum_{p\leq N}e^{2\pi i m p^{1/p}}$$
tends to zero as $N\to\infty$, $\forall m\neq 0$, where we are denoting the prime-counting function by $\pi(x)$ and $p$ the sequence of prime numbers in increasing order. 
I take the definition following the teacher in this video of the more high quality, to do the calculation $$L(s,\psi_m)=\prod_{p}\left(1-\frac{p^{\frac{2\pi i m}{p}}}{p^s}\right)^{-1}=\zeta(s-\frac{2\pi i m}{p}).$$
Now I believe that it is required to say that one has a pole at $s=1+\frac{2\pi i m}{p}$ (notice that I believe that I have a sequence of poles, Murty's example only depends on $m$), and say that I can use the theorem to show that our sum $\rightarrow 0$ as $N\to\infty$, thus our sequence is equidistributed in the (compact) group $\mathbb{R}/\mathbb{Z}$.
Can you explain if it is feasible such calculations or do it? I don't understand well if it is possible to  apply the theorem for my example, if you can state in details how one can use the theorem, then it is the best, and this example will be here as a reference. Thanks in advance.

As $n$ goes to infinity, $n^{1/n}$ approaches $1$ from above. In particular the fractional part of $n^{1/n}$ approaches $0$, so the $p^{1/p}$ are not equidistributed modulo $1$
