Question about convergence of series using comparison test 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?
 A: Yes, the series in question does diverge. What your professor used is known as the Limit Comparison Test. The proof of the test can be found in the Wikipedia article.
A non-rigorous, intuitive way to think about this is that if one series diverges to infinity, while the other converges to a real positive number, then you know that taking the limit of $\frac{a_n}{b_n}$ will result in either $\infty$ in the numerator or in the denominator and thus a limit of either $0$ or $\infty$.
A: What we are using is limit comparison test and since
$$\frac{\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2}}{\frac{1}{n}}=\frac{\sin{\frac{2}{\sqrt{n}}}}{\frac{2}{\sqrt{n}}}\frac{2}{3+2\frac{\sqrt n}n} \to \frac23$$
as you noticed the series diverges.
Note that limit comparison test also works for the two limiting cases

*

*$b_n$ converges and $\frac{a_n}{b_n} \to 0 \implies a_n$ converges

*$b_n$ diverges and $\frac{a_n}{b_n} \to \infty \implies a_n$ diverges

As an alternative by direct comparison test, using that $\sin x\ge x-\frac16 x^3$ we have
$$\frac{\sin{\frac{2}{\sqrt{n}}}}{3\sqrt{n}+2} \ge \frac{\frac{2}{\sqrt{n}}-\frac{4}{3n\sqrt{n}}}{3\sqrt{n}+2}=\frac{2-\frac{4}{3n}}{3n+2\sqrt{n}}\ge \frac{\frac23}{3n+3n}=\frac19\frac1n$$
