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Possible Duplicate:
Prove the divergence of the sequence $(\sin(n))_{n=1}^\infty$.

How can I show that the sequence $ a_n = \sin(n) $ is divergent? I tried to show that $\sin(n+1) - \sin(n)$ get always larger than some constant, but I did not succeeded.

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    See [here](http://math.stackexchange.com/questions/238997/prove-the-divergence-of-the-sequence-sinn-n-1-infty).2012-12-17

2 Answers 2

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You (presumably) know, or can prove, the following facts:

  • $\sin x > \dfrac{1}{2}$ whenever $2\pi n + \dfrac{\pi}{6} < x < 2\pi n + \dfrac{5\pi}{6}$, $n \in \mathbb{Z}$;
  • $\sin x < -\dfrac{1}{2}$ whenever $2\pi n - \dfrac{5\pi}{6} < x < 2\pi n - \dfrac{\pi}{6}$, $n \in \mathbb{Z}$;
  • For each $n$, each of the intervals mentioned above contains an integer.

So it is possible to pick an increasing sequence of integers, $b_n$ say, with $\sin (b_n) < -\dfrac{1}{2}$ when $n$ is odd and $\sin(b_n) > \dfrac{1}{2}$ when $n$ is even. What does this tell you?

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Your $(a_n)$ is bounded, but it doesn't converge. To see this let $k_n=[2n\pi+\frac{\pi}{2}]$ and $l_n=[2n\pi+\frac{3\pi}{2}]$ where $[x]$ denotes the integer part. Then, $a_{k_n}> 0$ while $a_{l_n}<0$. Two subsequences have different limits and so, $(a_n)$ diverges

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    @Nameless: It's certainly possible that that's true, but you should demonstrate that it's the case. It's not immediately obvious that you can't choose $n$ so that $\sin (k_n)$ is arbitrarily close to zero, for instance.2012-12-17