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According to theorem every bounded sequence in $R$ has a convergent subsequence.Can anybody make a convergent subsequence of $\sin{n}$.

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  • $\begingroup$ I can give you an algorithm for finding a convergent subsequence. $\endgroup$ – Matt Samuel Dec 18 '15 at 7:42
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Here's a semi-explicit one. The convergents of the continued fraction of $\pi$ are a sequence of rational approximations $p_n/q_n$ with $p_n$, $q_n$ positive integers tending to $\infty$, $$ \left| \pi - \dfrac{p_n}{q_n}\right| < \dfrac{1}{q_n^2}$$ and thus $$|\sin(p_n)| = |\sin(p_n - \pi q_n)| \le |p_n - \pi q_n| < 1/q_n \to 0 \ \text{as}\ n \to \infty$$

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Start with $a_1=1$, giving you $\sin(1)$. Now define $a_{n+1}$ to be the smallest integer $k$ such that

  • $k>a_n$
  • $0<\sin(k)<\sin(a_n)$

The second condition can be met since $\{\sin(n)\}$ is dense. Then $\{\sin(a_1),\sin(a_2),\ldots\}$ is such a sequence. It is decreasing and bounded below, so it converges to something (maybe $0$).

This is a recursive construction, and perhaps you are hoping for an explicitly defined sequence instead with some formula for the $n$th term. If that's what you are after, you could modify the question.

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