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I need to prove that the sequence $\{a \log n \}$ is NOT equidistributed for any $a \in \mathbb{R}$.

Now, I think that Weyl's criterion will be a good idea to use here. So, I need to show that $$\frac{1}{N} \sum_{n=1}^{N} e^{2 \pi i k (a \log n)} \rightarrow 0 $$ as $N \rightarrow \infty$ for any $k \in \mathbb{Z} - \{0\}$ doesn't hold.

I have two problems here. Firstly, I don't know whether the base of $\log$ is $10$ or $e$. So, this can be tried for both bases.

Secondly, if I assume that the base is $e$, then I have $$ e^{\log n^{a 2 \pi i k }}$$ inside the summation. This is equal to $n^{a 2 \pi i k}. $ So, I am left with $$\frac{1}{N} \sum_{n=1}^{N} n^{ 2 \pi i k a} $$ Now suppose I take a negative $k$ and positive $a$ or vice-versa, then wouldn't this series converges to $0$ as $N \rightarrow \infty$ ?

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  • $\begingroup$ You can prove this without exponential sums. The idea is that the function grows really, really slowly. You can assume the sequence is equidistributed and use this to get a contradiction. $\endgroup$ – Vik78 Oct 2 '16 at 4:45
  • $\begingroup$ Sorry but I don't get the idea of this function being really slow? You are talking about the whole summation function, right? $\endgroup$ – Dark_Knight Oct 2 '16 at 4:54
  • $\begingroup$ No, i'm talking about the logarithm function. Just say you look at the elements of the sequence out to index $n$, such that the proportion of elements up to $n$ within an interval $I \subset [0, 1)$ is within $\epsilon$ of the length of $I$, $a_n$ is not in $I$ and $a_{n+1}$ is. If the sequence next exits $I$ up at index $m$, you should be able to choose $n$ large enough such that the proportion of elements in $I$ at index $m$ to the total is no longer within $\epsilon$ of the length of $I$. This is because the log grows so slowly: you can write log $n \le$ log$(n + k) \le$ log $n$ $+ k/n$. $\endgroup$ – Vik78 Oct 2 '16 at 8:28
  • $\begingroup$ Indeed you can see from the above bounds that $m$ should be asymptotically at least $n |I|$. $\endgroup$ – Vik78 Oct 2 '16 at 8:42
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    $\begingroup$ I don't even know the equidistribution criterion, this is just how I proved it. If $r(k)$ is the number of times $(a_n)$ falls in $I$ before the $k$th index, find $k$ such that $r(k)/k$ is within $\epsilon$ of the length $|I|$ of $I$ and $a_k$ is in $I$. Let $m$ be the next index such that $a_m$ is not in $I$. As I explained above, as $k \to \infty$ we have $m \ge k + |I|k$. Therefore $r(m)/m \ge \frac{r(k) + |I|k}{k + |I|k} = \frac{r(k)}{k} + \frac{|I|}{1 + |I|}$. If you choose $|I|$ and $\epsilon$ small enough this is about $2|I|$, so $r(n)/n$ cannot converge to the length of $I$. $\endgroup$ – Vik78 Oct 3 '16 at 21:30
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Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$.

For $f(x)=e^{ic\log x}$ with $c=2\pi ka$, we have $f'(x)=icf(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{ic\log N}=I_N+F_N+R_N, \\I_N=\frac1N\left.\frac{e^{(1+ic)\log x}}{1+ic}\right|_{x=1}^{x=N}=\frac{e^{ic\log N}-1/N}{1+ic}, \\F_N=\frac{1+e^{ic\log N}}{2N},\quad|R_N|\leqslant\frac{|c|}{N}\int_1^N\frac12\frac{dx}{x}=\frac{|c|}{2}\frac{\log N}{N}.$$

As $N\to\infty$, clearly $F_N\to 0$ and $R_N\to 0$ but $I_N\not\to 0$. Hence, $W_N\not\to 0$.

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