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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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    No. By diverges, it doesn't mean that the sequence is unbounded. It means that the sequence does not converge to a real number.2012-12-17
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    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?2012-12-17
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    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?2012-12-17
  • 0
    @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.2012-12-17
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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

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