# 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.

• If you can see the video seach the channel mathsciencechannel, and after the label videos after few pages you find it. – user243301 Jul 19 '16 at 18:48
• Are you interested in the fractional part of $p^{1/p}$? This is not equidistributed, the fractional parts approach $0$. – André Nicolas Jul 19 '16 at 18:52
• For all $m \neq 0$ the limit is simply $e^{2 \pi i m} \neq 0$, so the answer is no. This can be easily proved using the fact that $p^{1/p} \to 1$. – Crostul Jul 19 '16 at 18:52
• No I only was interested to do an exercise and encourage to me to study more mathematics. Thanks, then has meaning the question? Thanks @AndréNicolas – user243301 Jul 19 '16 at 18:53
• Then you are welcome to add a detailed answer, because I try do this as first example. I don't understand well the theory, Then you can add an answer to say that isn't equidistributed. Very thanks much @Crostul – user243301 Jul 19 '16 at 18:55

Using the result of basic calculus in this link, we can say $$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=1}^n e^{2 \pi i m p_k^{1/p_k}}= \lim_{p \to +\infty} e^{2 \pi i m p^{1/p}}$$ , where $p_i$ denotes the $i$-th prime number. Now, it is well known that $$\lim_{n \to +\infty}n^{1/n}=1$$ so that $$\lim_{p \to +\infty} e^{2 \pi i m p^{1/p}}=e^{2 \pi i m \cdot 1}=e^{2 \pi i m} = e^{\mbox{something}} \neq 0$$
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 $1$