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Given $a_n = \sin(\frac{n\pi}{4})$, I need to prove that a there is no limit for this sequence.

I think proof by contradiction is a good method. Suppose not, that is $a_n$ has a limit equals to L. Since $-1 < a_n < 1$, $a_n$ is a bounded sequence. Thus, according to Bolchano-Weishtrass theorem, any bounded sequence has a convergent subsequence. How can I find all subsequences of $a_n$ to show that $a_n$ does not have a limit?

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We don't need Bolzano-Weirstrass theorem, indeed for the unicity of the limit theorem, it suffices to show that at least 2 subsequences exist with different limit.

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You can prove that $(u_n)$ has at least two different adherence values. Indeed, $$\forall n\in\mathbb{N},\,u_{8n}=0$$ and $$ \forall n\in\mathbb{N},\,u_{8n+2}=1 $$

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  • $\begingroup$ If I show that the subsequence of $a_n$ has two different values, what does this mean? Does it mean two subsequences do not have the same limit and then there is a contradiction? $\endgroup$ – Math learner Oct 4 '19 at 21:07
  • $\begingroup$ Absolutely, if $(a_n)$ converges toward $\ell$, then every subsequence of $(a_n)$ also converges toward $\ell$. $\endgroup$ – Tuvasbien Oct 4 '19 at 21:26

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