calculus - Show by comparison that $sumlimits_{n=1}^infty sin(frac{1}{n})$ diverges?

By using the integral test, I know that $$\sum\limits_{n=1}^\infty \sin\left(\frac{1}{n}\right)$$ diverges. However, how would I show that the series diverges using the limit comparison test? Would I simply let $\sum\limits_{n=1}^\infty a_n = \sum\limits_{n=1}^\infty b_n = \sum\limits_{n=1}^\infty \sin\left(\frac{1}{n}\right)$ 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)?

Answer

Notice $x_n = \frac{1}{n}$, then $\sum \frac{1}{n}$, the harmonic series, we all know is divergent. Now,

$$\lim \frac{ \sin (\frac{1}{n})}{\frac{1}{n}} =_{t = \frac{1}{n}} \lim_{t \to 0} \frac{ \sin t}{t} = 1$$

The result now follows by the limit comparison test. :)