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,
Best Answer
Note it suffices we check this for $x\geqslant 0$. We can use that $\sin{x}n^{-2}\leqslant {x}{n^{-2}}$ and that if $x>0$ is fixed and $n>N$ large enough, $\sin({x}{n^{-2}})\geqslant 0$ (since $\sin$ is positive on $(0,\pi/2)$). Thus, by comparison, $$0\leqslant \sum_{n>N}\sin(xn^{-2})\leqslant x\sum_{n>N} n^{-2}<\infty$$
This in fact shows convergence is uniform in every compact subset of $\Bbb R$. You can also simply use $|\sin y|\leqslant |y|$ to avoid the argument that the summands are eventually positive.