$\sum_{n=1}^\infty\frac{(-1)^n\sin(3n)}{\sqrt{n^2+1}}$
So I tried the absolute convergence: $ \frac{\lvert \sin(3n)\rvert}{\sqrt{n^2+1}} \ge \frac{\sin^2(3n)}{n} \sim \frac{1}{2n} - \frac{\cos 6n}{2n} $.$ \frac{1}{2n} $ doesn't converge, while $ \frac{\cos 6n}{2n} $ converge, so the whole series doesn't converge absolutely. But I don't know how to prove, if it converge or not at all. I stacked here: $ \frac{(-1)^n\sin(3n)}{\sqrt{n^2+1}} = \frac{\sin(\frac{n\pi}{2})\sin(3n)}{\sqrt{n^2+1}} = \frac{\cos n(\frac{\pi}{2}-3) - \cos n(\frac{\pi}{2}+3)}{2\sqrt{n^2+1}} $. Help with the next step or explain a different solution please.
This is a consequence of Abel's Lemma:
If $\{b_n\}$ is non-negative, decreasing and tends to zero, and the partial sums of $\{a_n\}$ are bounded, then the series $\sum a_nb_n$ converges.
Here:
$$ a_n=(-1)^n\sin(3n), \quad b_n=\frac{1}{\sqrt{n^2+1}}. $$