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How to prove that $$\lim_{n\to\infty}{\sin{100n}}$$ doesn't exist?

Some possible approaches:

  1. It would be enough to find two subsequences $n_{k}$ that converge to two different numbers. But it's not clear how to find $n_k$ so that $\sin 100n_k$ converge.

  2. Show that $\sin (100(n+1))-\sin 100n$ does not approach $0$. This is not obvious, either.

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    $\begingroup$ I don't understand the downvote/close vote for this question. Carefully writing out the proof of the divergence of a series like $(\sin(100n))$ for the first time is not trivial. $\endgroup$ – Simon S Nov 28 '14 at 19:14
  • $\begingroup$ Can you find two subsequences that converge to different limits? $\endgroup$ – Mike Pierce Nov 28 '14 at 19:14
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    $\begingroup$ @GDumphart Bolzano Weierstass theorem. $\endgroup$ – snulty Nov 28 '14 at 19:21
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    $\begingroup$ @snulty Ok sorry, my bad :D $\endgroup$ – GDumphart Nov 28 '14 at 19:22
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    $\begingroup$ @GDumphart I disagree. Can't we find some (probably not pretty) sequence of natural numbers $s_n$ such that the sequence $|100s_n\;(\mathrm{mod}\;2\pi)|$ approaches $\frac{\pi}{2}$? This would be the subsequence that approaches $1$, and we can similarly find a subsequence that approaches $-1$. $\endgroup$ – Mike Pierce Nov 28 '14 at 19:28
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Hint: You can use the fact that $$\sin m-\sin n=2\sin\frac{m-n}{2}\cos\frac{m+n}{2}.$$

Assume for instance that $m=n+2$ and let $n\to \infty$.

Added in Edit: If the limit existed, then of course the LHS of the above equation would be zero in limit, and so would be the RHS. This would imply that $\cos n\to 0$ which is a contradiction.

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    $\begingroup$ How does this help me? I don't see it. $\endgroup$ – user196176 Nov 28 '14 at 19:47
  • $\begingroup$ @user196176 I added some more explanation to my answer. $\endgroup$ – EPS Nov 28 '14 at 20:15
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    $\begingroup$ I think with respect to the posed problem, showing $\cos n \not\rightarrow 0$ is quite similar to what's being asked. $\endgroup$ – snulty Nov 28 '14 at 20:18
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    $\begingroup$ But what I think I mean is that, how do you know that for $k \equiv 1\pmod{2}$, and with $k$ and $n$ large, $k\pi/2$ and $n$ don't become arbitrarily close. I think that step is worth mentioning? $\endgroup$ – snulty Nov 28 '14 at 20:30
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    $\begingroup$ But how do I connect this with $\sin{100n}$? $\endgroup$ – user196176 Nov 28 '14 at 20:30
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For any $L\in\mathbb{R}$ and $N\in\mathbb{Z}_+$, you can find some $n>N$ such that $|\sin(100n) - L| > \frac{1}{2}$. As a construction, suppose $L \leq 0$ and find a value of $n$ that satisfies $\frac{\pi}{6} < 100n + m2\pi < \frac{5\pi}{6}$ for some $m \in \mathbb{Z}$. (If you really need a construction, you can write down a formula for $n$ with a mess of floor functions.) Then $\sin(100n) > \frac{1}{2}$, so $|\sin(100n) - L| > \frac{1}{2}$. Similarily for $L \geq 0$.

Thus not all values of $\epsilon > 0$ can satisfy the limit requirement $|\sin(100n) - L| < \epsilon$, so the limit does not exist.

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  • $\begingroup$ But how do you actually prove that statement? $\endgroup$ – user196176 Nov 28 '14 at 19:45
  • $\begingroup$ I've added a construction of $n$. $\endgroup$ – Vortico Nov 28 '14 at 19:57

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