# Show that a sequence does not admita limit using BW theorem

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?

You can prove that $$(u_n)$$ has at least two different adherence values. Indeed, $$\forall n\in\mathbb{N},\,u_{8n}=0$$ and $$\forall n\in\mathbb{N},\,u_{8n+2}=1$$
• 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? – Math learner Oct 4 '19 at 21:07
• Absolutely, if $(a_n)$ converges toward $\ell$, then every subsequence of $(a_n)$ also converges toward $\ell$. – Tuvasbien Oct 4 '19 at 21:26