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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.

user196176
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    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. – Simon S Nov 28 '14 at 19:14
  • Can you find two subsequences that converge to different limits? – Mike Pierce Nov 28 '14 at 19:14
  • You can use that the sine always alternates between $-1$ and $1$ and thus will always leave a neighborhood of e.g. $\varepsilon = \frac{1}{2}$ for some higher $n$. – GDumphart Nov 28 '14 at 19:15
  • @mapierce271 No. It's the limit of the sequence. ($n\in\mathbb N$). – user196176 Nov 28 '14 at 19:16
  • @GDumphart but how do I show it? – user196176 Nov 28 '14 at 19:17
  • @mapierce271 I think a sine-sequence with integer angular frequency $\omega = 100$ doesn´t have any convergent subsequence. – GDumphart Nov 28 '14 at 19:17
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    @GDumphart Bolzano Weierstass theorem. – snulty Nov 28 '14 at 19:21
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    @snulty Ok sorry, my bad :D – GDumphart Nov 28 '14 at 19:22
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    @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 $-1$. – Mike Pierce Nov 28 '14 at 19:28
  • @mapierce271 Yes sure, my false statement got already smashed by snulty. – GDumphart Nov 28 '14 at 19:30
  • http://math.stackexchange.com/questions/1042823/convergence-test-of-the-series – snulty Nov 28 '14 at 21:04
  • See also http://math.stackexchange.com/questions/27218/prove-that-the-limit-of-sin-n-as-n-rightarrow-infty-does-not-exist (and other questions linked to that one). – Martin Sleziak Feb 02 '15 at 11:06

2 Answers2

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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.

EPS
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    How does this help me? I don't see it. – user196176 Nov 28 '14 at 19:47
  • @user196176 I added some more explanation to my answer. – EPS Nov 28 '14 at 20:15
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    I think with respect to the posed problem, showing $\cos n \not\rightarrow 0$ is quite similar to what's being asked. – snulty Nov 28 '14 at 20:18
  • @snulty I have to disagree! We know how to solve $\cos x=0$, don't we? – EPS Nov 28 '14 at 20:19
  • I think I'm too tired to realize if I'm just being silly. – snulty Nov 28 '14 at 20:27
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    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? – snulty Nov 28 '14 at 20:30
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    But how do I connect this with $\sin{100n}$? – user196176 Nov 28 '14 at 20:30
  • @user196176 replace $m$ and $n$ with $100m$ and $100n$ – snulty Nov 28 '14 at 20:32
  • @snulty What I have written is just a hint that after filling out minor details becomes a full solution. As for your question, assume that $n$ is very close' to $k\pi/2$ for an odd $k$ (whatever very close might mean). Can $n+1$ be alsovery close' to some $k'\pi/2$? – EPS Nov 28 '14 at 22:44
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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.

Vortico
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