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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

Martin Sleziak
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TKM
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  • Switching the sums is always possible ; a finite sum of limits is the limit of the finite sums (the limit I am speaking of here is the series, not your limit indexed by $N$), assuming the series converges (in this case this is the series of the exponential so there is absolutely no problem). The big issue is really 2 ; limits and series don't commute in general, and it is the issue most of the time in analysis, so... – Patrick Da Silva Nov 02 '14 at 17:26
  • what is the condition for commuting? That they both converge? If so does that follow, since they equal zero? – TKM Nov 02 '14 at 20:39
  • There are no general conditions for this, there are only particular cases! In the general case you use your brain and try to pull off tricks to prove convergence. You can bound the error term, use things like Fubini's theorem, etc... and this is not easy in general. – Patrick Da Silva Nov 02 '14 at 21:58

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