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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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marked as duplicate by David Mitra, Clive Newstead, Marc van Leeuwen, QiL, froggie Dec 17 '12 at 16:57

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No. By diverges, it doesn't mean that the sequence is unbounded. It means that the sequence does not converge to a real number. –  Gautam Shenoy Dec 17 '12 at 15:50
As $-1\leq \sin(x)\leq 1 $ you wont succed showing that $\sin(n+1)-\sin(n)$ gets bigger than a constant, but if $a_n$ would converge, there would be a value $b=\lim_{n\rightarrow \infty} \sin(n)$. What can you say about that? –  sonystarmap Dec 17 '12 at 15:52
because it is bound, if it converges the limit equals $\lim \sup a_n$ and $\lim \inf a_n$, and these two values are not equal, but how can i show that $\lim \sup a_n = 1$ for n being integers? –  Stefan Dec 17 '12 at 15:59
@stefan, does $b_n=(-1)^n$ converge? Because it is bounded, too. Boundedness only implies that some sub-sequence converges, not that the entire sequence converges. –  Thomas Andrews Dec 17 '12 at 16:00
See here. –  David Mitra Dec 17 '12 at 16:04

2 Answers 2

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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Pressumably, since he is talking about a sequence, $k_n$ needs to be an integer. After all, your argument works for $b_n=\sin 2\pi n$, too. –  Thomas Andrews Dec 17 '12 at 15:57
@ThomasAndrews I think I fixed it. –  Nameless Dec 17 '12 at 16:17
@Nameless: How do we know they don't both converge to zero? [For instance, the sequence $a_n = \frac{(-1)^n}{n}$ has $a_n>0$ for $n$ even and $a_n<0$ for $n$ odd, and yet it converges.] –  Clive Newstead Dec 17 '12 at 16:25
@CliveNewstead I think we can bound $k_n$ so that $a_{k_n}$ is constantly above something positive (using monotonicity of $\sin$ in the intervals we are interested in) –  Nameless Dec 17 '12 at 16:27
@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. –  Clive Newstead Dec 17 '12 at 16:30

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