I would like you to take a look at the following function-series. This is a problem I came across in Real Analysis some months ago and I had left it aside, but then I forgot about it. Note that Weierstrass's criterion is not applicable here and I can't seem to find a factorization that would make either Abel's or Dirichlet's criterion applicable.
$\displaystyle{s(x)=\sum_{n=1}^{\infty}2^n\sin\left(\frac{1}{3^nx}\right)}$, for $x \in (0,+\infty)$. By the $\sin(x)<x$ inequality, it is obvious that it converges pointwise. It is also quite obvious that it converges uniformly in any $[\alpha,+\infty)$, where $\alpha>0$. What about $(0,+\infty)$ though?
Let $f_n(x) = 2^n \sin \frac{1}{3^nx}$. If the series were uniformly convergent on $(0,+\infty)$, then we'd have uniform convergence of $(f_n)$ to $0$. But
$$f_n(3^{-n}) = 2^n \sin 1$$
is unbounded.