Prove there is a sequence of increasing positive integers $n_i$ s.t. the limit of $\lim_{{n_i} \to \infty} \sin(n_i)$ exists Here is my question: Prove there is a sequence of increasing positive integers $n_i$ such that the limit of $\sin(n_i)$ exists as $i \to \infty $. The problem with this question is... I don't know how to start it! Doesn't the sine function fluctuate between -1 and 1? How would a limit exist if it keeps going back and forth? Thank you for any help you give me! :)
 A: Take $a_n = \sin (n)$ then since $[-1,1] \supset \bigcup_{n=0}^{\infty} a_n$ and $[0,1]$ is compact these points have an accumulation point in $[-1,1]$.
Hence you have found $a_{k_n} \rightarrow A$, to find an increasing sequence of indeces use the fact when you have $\phi :\mathbb{N}\rightarrow \mathbb{N}$ then also $a_{\phi(k_n)}\rightarrow A$.
A: A consequence of Dirichlet's approximation theorem is that there exists infinitely many positive integers $n_k$ such that $\left|\pi-\dfrac{p_k}{n_k}\right|\leqslant\dfrac1{n_k^2}$ for some integer $p_k$. Then $|n_k\pi-p_k|\leqslant\dfrac1{n_k}$ hence $|\sin(p_k)|\leqslant\dfrac1{n_k}$. In particular, $\sin(p_k)\to0$ when $k\to\infty$.
A: The sequence whose $n$-th term is $\sin n$ is a bounded (by $\pm1$) sequence of real numbers.  By the Bolzano-Weierstrass Theorem, it has a convergent subsequence.  (You should certainly make your peace with this theorem before worrying about more advanced things like Dirichlet's Theorem or Thue-Roth-Siegel.)
A: Hint: in the Reals, continuity is equivalent to sequential continuity, i.e., if $x_n \rightarrow x $ , then $f(x_n) \rightarrow f(x) $ . And $\sin x$ is continuous.
Edit: What I meant is we can use a sequence to approximate a multiple of $\pi$, and then use sequential continuity. This follows from Roth-Thue-Siegel. Equivalently, by Weil's equidistribution theorem http://en.wikipedia.org/wiki/Equidistribution_theorem  on {$n\pi$}, the sequence $n\pi$  is equidistributed $mod1$ , so that it will be indefinitely-close to an integer infinitely often, say at $n_1,n_2,....$. Take these terms $n_i$ of the sequence that are within $<\epsilon$ of an integer for the actual sequence $n_i$. Then $n_i -k\pi <\epsilon$ , so that $Sinn_i \rightarrow Sink\pi=0$.
This is the specific argument, which apparently I have to provide, but others do not: the sequence $n \pi; n=1,2,3,...$ considered $Mod1$ , meaning we consider $nx$ minus the greates integer less than $nx$ , i.e., the decimal part of $nx$. These decimal parts are dense in $[0,1]$. This means, in particular, that for any $\epsilon/n$ , there is an integer $m$ with $m\pi-k < \epsilon $ for some integer $k$. Since $n\pi$ is dense ($Mod1$ in $[0,1]$) , we can find an integer $m_2>m$ , so that $|m_2\pi -k_2| <\epsilon/2$.  The same can be done for any $\epsilon/n$ , i.e., we can find an increasing sequence of integers $m_i$ , so that $|m_ipi -k_i < \epsilon/I $ . The sequence $k_i$ is indefinitely -close to integer multiples of $\pi$. Then $sink_i \rightarrow sin\pi=0$. 
