Convergence of series of $\sin(x/n^2)$ For the convergence of series of $\sin\left(\frac{x}{n^2}\right)$, is it enough to say that since, for large $n$, $$a_n:= \sin\left(\frac{x}{n^2}\right) \approx b_n:= \frac{x}{n^2},$$ so that $\displaystyle\lim_{n\to\infty} \dfrac{a_n}{b_n} \ \text{ exists}$, and by the limit comparison test, series $a_n$ and series $b_n$ converge or diverge together - and since series $b_n$ is a convergent $p$-series for all $x$, series an is convergent for all $x$, too?  Or am I missing something?
Thanks,

It is valid for series with positive terms. However, you can use it to prove the series is absolutely convergent: $ \biggl\lvert\sin\biggl(\dfrac{x}{n^2}\biggr)\biggr\rvert \sim\dfrac{\lvert x\rvert}{n^2} $, hence it converges.
