Is $\{(\frac{3}{2})^{n}\}$ for $n\in \mathbb{N}$ dense in [0,1] (open question). By $\{(\frac{3}{2})^{n}\}$ I mean the fractional part of $(\frac{3}{2})^{n}$.
A more general question is: Is $\{(\frac{3}{2})^{n}\}_{\mathbb{N}}$ is equidistributed in [0,1]?If so, Then density follows.
I am just curious if the approach below is true i.e. if steps (1) and (2) are true.
Asume $\{(\frac{3}{2})^{n}\}_{\mathbb{N}}$ is equidistributed i.e. by Weyl's theorem that
$0=lim_{N\to \infty}\frac{1}{N}\sum_{k=1}^{N}e^{2\pi i (\frac{3}{2})^{k}}=$
Then we do Taylor expansion.
$=lim_{N\to \infty}\frac{1}{N}\sum_{k=1}^{N}\sum_{m=1}^{\infty}\frac{(2\pi i (\frac{3}{2})^{k})^{m}}{m!}=$
(1)Can we switch the sums? If we can, then we get
$=lim_{N\to \infty}\frac{1}{N}\sum_{m=1}^{\infty}\frac{(2\pi i )^{m}}{m!}\sum_{k=1}^{N}(\frac{3}{2})^{km}=$
$=lim_{N\to \infty}\frac{1}{N}\sum_{m=1}^{\infty}\frac{(2\pi i )^{m}}{m!}\frac{1-(\frac{3}{2})^{m(N+1)}}{1-(\frac{3}{2})^{m}}=$
$=lim_{N\to \infty}\frac{1}{N}\sum_{m=1}^{\infty}\frac{(2\pi i )^{m}}{m!}\frac{1-(\frac{3}{2})^{m(N+1)}}{1-(\frac{3}{2})^{m}}.$
(2)Can we take the limit of the inside? If we can, then:
$lim_{N\to \infty}\frac{1-(\frac{3}{2})^{m(N+1)}}{N}=\infty$ because exponential grows faster than polymonial.
Thus we get a contradiction.
I think (1) is not fixable because the summand is oscillatory. What do you think?
Question: Are steps (1), (2) true?
Thanks