Consider the following series on the set $A=\{x\in\mathbb{R}\mid x>0\}$: $$\sum_{n=1}^{+\infty}2^n\sin\frac{1}{3^nx}$$
Does the series converge uniformly on $A$ or some subsets?
The idea is to obeserve that:
$$\sum_{n=1}^{+\infty}2^n\sin\frac{1}{3^nx}=\sum_{n=1}^{+\infty}\frac{2^n}{3^nx} \Big(3^nx\sin\frac{1}{3^nx}\Big)$$
If we were able to say that $\sin\frac{1}{3^nx}<\frac{1}{3^nx}$ for all $x$ then the exercise ends by Weiestrass criterion, but such inequality needs not hold for every $x\in A$. How do we proceed now?
It converges uniformly in every set of the form $$ A=[a,\infty), $$ where $a>0$.
Proof. Using the fact that $$ |\sin x|\le |x|, $$ we have that for $x>0$ $$ \left|\sin \frac{1}{x}\right|\le\frac{1}{x} $$ So if $x\in [a,\infty)$, then we have $$ \left|2^n\sin\frac{1}{3^nx}\right|\le\frac{2^n}{3^nx}\le \frac{2^n}{3^na}, $$ and $\sum \frac{2^n}{3^na}<\infty,$ and hence the series converges absolutely and uniformly in $[a,\infty)$.
On the other hand the series DOES NOT converge uniformly in $(0,\infty)$. To see this, consider $$ x_n=\frac{2}{3^n\pi}, $$ and $$ s_n(x)=\sum_{k=1}^n 2^k\sin\frac{1}{3^kx}. $$ Then $s_n(x_n)-s_{n-1}(x_n)=2^n$, and hence the sequence $\{s_n(x)\}$ does not converge uniformly.