# Using the (limit) comparison test to test $\sum\limits_{n=1}^\infty\sin\frac1n$ for con-/divergence

Problem:

Use the comparison test, or limit comparison test, to see if $$\sum\limits_{n=1}^\infty\sin\frac1n$$ converges or diverges.

My attempt:

Sadly empty. So far, I've only dealt with sums where the terms are polynomial/polynomial, but for this one I'm stuck.

Any help appreciated!

Since $0 < 1/n \le 1 < \pi/2$ for all $n\in \Bbb N$, $\sin \frac{1}{n} > 0$ for all $n \in \Bbb N$. Now
$$\lim_{n\to \infty} n\sin \frac{1}{n} = \lim_{x\to 0} \frac{\sin x}{x} = 1$$
and $\sum_{n = 1}^\infty \frac{1}{n}$ diverges, so by the limit comparison test, $\sum_{n=1}^\infty \sin \frac{1}{n}$ diverges.
$$\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}}= \color{red}1$$