# How would we show by comparison that $\sum\limits_{i=1}^\infty \sin(\frac{1}{n})$ diverges?

By using the integral test, I know that $\sum\limits_{i=1}^\infty \sin\left(\frac{1}{n}\right)$ diverges. However, how would I show that the series diverges using the limit comparison test? Would I simply let $\sum\limits_{i=1}^\infty a_n = \sum\limits_{i=1}^\infty b_n = \sum\limits_{i=1}^\infty \sin\left(\frac{1}{n}\right)$ and then take $\displaystyle \lim_{n \rightarrow \infty}\frac{a_n}{b_n}$ to show the series diverges (assuming the limit converges to some nonnegative, finite value)?

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For Limit Comparison, compare with $\sum \frac{1}{n}$. – André Nicolas Feb 4 '14 at 0:47
Why not do straight comparison with $\sum \frac{1}{n}$? – Chris Leary Feb 4 '14 at 1:10
Disregard my comment. What I was thinking was nonsense. – Chris Leary Feb 4 '14 at 1:17

Notice $x_n = \frac{1}{n}$, then $\sum \frac{1}{n}$, the harmonic series, we all know is divergent. Now,
$$\lim \frac{ \sin (\frac{1}{n})}{\frac{1}{n}} =_{t = \frac{1}{n}} \lim_{t \to 0} \frac{ \sin t}{t} = 1$$