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**Pinkk** I have to say Plato, I did think that was a tiny bit arrogant to ask. It should be obvious, but better to ask than to just assume!

If you wanna formalize why there can be no convergent subsequence, we can assume to the contrary. Let [tex]s_{n} = 2\pi n[/math[ and assume that there is a subsequence $\displaystyle (s_{n_{k}})$ such that given $\displaystyle \epsilon > 0$, there exists $\displaystyle N$ such that $\displaystyle n_{k} > N$ implies $\displaystyle |s_{n_{k}} - a| < \epsilon$, where $\displaystyle a$ is the finite number that we suppose the subsequence will converge to. And so this means for all $\displaystyle n_{k} > N$, $\displaystyle s_{n_{k}} < \epsilon + a$ (assume that $\displaystyle a > 0$ because if it wasn't, then clearly after some point the sequence would have to be negative). But since $\displaystyle \epsilon + a > 0$, we know for *all* $\displaystyle n > \frac{\epsilon + a}{2\pi}$, $\displaystyle 2\pi n > \epsilon + a$, and so rises the contradiction.