$$\sum\limits_{n=1}^{\infty} n^2 \sin(\frac{x}{n^4})$$ I am asked to study the pointwise and uniform convergence of the series.
For the pointwise requirement, it is convenient to study the asymptotically equivalent: $$\sum\limits_{n=1}^{\infty} n^2 \frac{x}{n^4}=\sum\limits_{n=1}^{\infty} \frac{x}{n^2}$$ Since it is an harmonic series of order two, the original series converges for all $x\in R$.
I wonder what is the appropriate way to study uniform convergence. I think that studying uniform convergence of an asymptotically equivalent series is not correct but I'd like to have some comments about it [for example, is it acceptable to use convergence radius of the asymptotically equivalent power series to discuss uniform convergence of a series?]
Using Weierstrass test and the inequality $\sin t < t$, I noticed that when considering set like $S=[a,b]$: $$\sup_{x \in S} |f_n(x)| \leq sup \frac{x}{n^2}=\frac{c}{n^2}$$ where $c=\max (|a|,|b|)$. Finally $\sum\limits_{n=1}^{\infty}\frac{c}{n^2}$ converges, so that the original series is uniformly convergent in $[a,b]$. Same argument doesn't apply when $S$ is not a limited set. Is this study of uniform convergence sufficiently exahustive? Can other uniform convergence sets exist? For example, I think that union of a finite number of disjoint intervals is still a solution. [Any compact set, actually].
Moreover, a general question: could a set of isolated points be the set of uniform convergence for a particular series?
Your series is uniformly convergent on any bounded subset. On all of $\mathbb{R}$, however, it is not uniformly convergent. One way to see this is: if it were uniformly convergent, the general term would converge to zero uniformly. But this is clearly not the case, since given any $n$, you can choose $x=n^4$ and, at that point, the general term is $n^2\sin 1$ which is greater than, say, $\sin 1$.