# Show that $(x_n)_n$ is convergent.

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.

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

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.

• Nice answer to a nice question. Will add to my list. Sep 9, 2021 at 15:46