I am wondering how to determine whether the following series converges using the comparison test:
$$\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$$
In class, my professor used the idea that $\frac{\sin x}{x}\to 1$ as $x$ gets large.
So, my professor said that since $$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}\to\frac{2}{3}$$by the comparison test, $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ converges. Can someone explain why this is true? I though that the comparison tests says that if $\lim_{n\to\infty} \frac{a_n}{b_n}\to L$ where $0<L<\infty$, then $a_n$ and $b_n$ follow the same convergence. So wouldn't this mean that $\sum_{n=1}^\infty \frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}$ diverges since $\frac{1}{n}$ diverges?