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$\textbf{Proof}$:

I need to show that $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$

Let $\epsilon = 2$, then if you take $n = 30$, we have that $a_{30} = sin(10) \approx -0.544 \ldots$

So we have that $ \left | sin(10) - 3 \right | \ge 2$

Therefore I have showed that, $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$, where $\epsilon = 2$ and $n = 30$.

I am wondering if this proof is right.

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    $\begingroup$ You don't need to calculate (even approximately) values of sine. Just use the fact that $\sin x \le 1$ for all $x$. $\endgroup$ – njguliyev Oct 31 '13 at 21:02
  • $\begingroup$ I will remove the approximations, they do look rather ugly. I was wondering if general argument I've made is correct though? $\endgroup$ – Michael Oct 31 '13 at 21:03
  • $\begingroup$ Ok Andre, I see my mistake. $\endgroup$ – Michael Oct 31 '13 at 21:05
  • $\begingroup$ You didn't make a general argument yet. You chose correct $\varepsilon$, but for any $N$ you must show that the inequality is true for some $n > N$. $\endgroup$ – njguliyev Oct 31 '13 at 21:06
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You have to find $\varepsilon>0$ such that, for all $N\in\mathbb N$, there exists $n\geq N$ such that $|a_n-3|\geq \varepsilon$. What you did will only work for $N\leq 30$, but, if for example $N=31$, you have to find an $n\geq 31$ such that $|a_n-3|\geq\varepsilon$.

Instead, you could do the following: given $N\in\mathbb N$, there exists $n\geq N$ such that $a_n=\sin k$ for some $k\in\mathbb N$. So, for that $n$, $|a_n-3|=|\sin k-3|\geq\left||3|-|\sin k|\right|\geq 3-|\sin k|\geq 2$.

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  • $\begingroup$ Ok I see my mistake, and your argument makes sense. I was jumping gun and finding specific n's, when it should be for a given n, as you've said. $\endgroup$ – Michael Oct 31 '13 at 21:12
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It suffices you show $\langle \sin n\rangle $ cannot converge to $3$. That is, if the sequence converged to $3$, then the subsequence $\sin 1,\sin 2,\ldots$ would converge to $3$. But for any $n$, we have that $\sin n \leqslant 1$. Thus, this is impossible. Note that if we replaced $3$ with $1$; things'd get much more delicate!

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  • $\begingroup$ Oh ok, this is very clever. Thanks for your response Pedro. $\endgroup$ – Michael Oct 31 '13 at 21:13
  • $\begingroup$ @Michael No problem. $\endgroup$ – Pedro Tamaroff Oct 31 '13 at 21:14

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