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Let $(x_n)_n$ be a bounded sequence. Let $a,b >0$ s.t. $\frac{a}{b}$ is an irrationnel. The sequences $(e^{iax_n})_n$ and $(e^{ibx_n})_n$ converge. Show that $(x_n)_n$ converges.

Starting with the fact that $(e^{iax_n})_n$ and $(e^{ibx_n})_n$ converge, we get by definition that:

Suppose that $(e^{iax_n})_n$ converges to $l_1$, then $\forall \epsilon_1 >0 \ \exists N_1>0$ s.t. $\forall n > N_1$: $|e^{iax_n}-l_1|< \epsilon_1$.
Suppose that $(e^{ibx_n})_n$ converges to $l_2$, then $\forall \epsilon_2 >0 \ \exists N_2>0$ s.t. $\forall n > N_2$: $|e^{ibx_n}-l_2|< \epsilon_2$.

From here I don't see how to show that $(x_n)_n$ is convergente.

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  • $\begingroup$ Intuitively I would say $\frac{a}{b}$? $\endgroup$
    – vitalmath
    Sep 9, 2021 at 15:35
  • $\begingroup$ Ohh.. never had a exercise like this before but I will work on it! Thanks $\endgroup$
    – vitalmath
    Sep 9, 2021 at 15:43
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    $\begingroup$ @TheSilverDoe Good point, it took me some time to understand that but I've got it now. $\endgroup$ Sep 9, 2021 at 16:04

1 Answer 1

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Let $x$ and $y$ be two limit points of $(x_n)$. Then, because the sequences $(e^{iax_n})$ and $(e^{ibx_n})$ both converge, you get $$e^{iax} = e^{iay} \quad \text{and} \quad e^{ibx} = e^{iby}$$

so $$ax = ay + 2k\pi \quad \text{and} \quad bx=by + 2l\pi$$

for some integers $k$ and $l$.

If $x \neq y$, you get $$a= \frac{2k\pi}{x-y} \quad \text{and} \quad b= \frac{2l\pi}{x-y} $$

which contradicts the fact that $a/b$ must be irrationnal. So $x=y$.

So you have a bounded sequence with only one limit point : it is necessarily a convergent sequence.

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    $\begingroup$ Nice answer to a nice question. Will add to my list. $\endgroup$ Sep 9, 2021 at 15:46

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