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So the limit as $n \rightarrow \infty$ of $\frac{2\sin(n)}{n}$ is $0$.

How do I prove this?

I say;

$$|\frac{2\sin(n)}{n} - 0 | = |\frac{2\sin(n)}{n}| < \epsilon$$ for which $n$'s ?

Not sure how to get rid of the absolute value bars, or even what to do next. First day with series and convergence in class!

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  • $\begingroup$ See my answer here for an explanation of the process. $\endgroup$ – Git Gud Sep 9 '14 at 15:52
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Use the fact that $\lvert \sin(x) \rvert \leq 1$ for all $x$. Then, you should be able to get a nice $N$ for a fixed $\varepsilon$.

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