Let $f(x)=\sum\limits_{n=1}^\infty e^{-nx}\cos(n^2x^2)$. It has been shown already that the series uniformly converges on $(0,\infty)$, and that $f(x)$ is continous on the same interval. The question is, is $f(x)$ differentiable?
I tried to use to following theorem, which is the only one we've studied about series's differentiation:
Let $\{f_n\}_{n=1}^\infty \in C^1(D, \mathbb{R})$, $\sum\limits_{n=1}^\infty f'_n(x)$ uniformly converges on $D$, and $\sum\limits_{n=1}^\infty f_n(c)$ converges for some $c \in D$, then $\sum\limits_{n=1}^\infty f_n(x)$ uniformly converges to $f(x)$ and $f'(x) = \sum\limits_{n=1}^\infty f'_n(x)$.
In this case, we get: $f_n' = -ne^{-nx}(\cos(n^2x^2) + 2n^2x\sin(n^2x^2)$. For any $n>1$, $f_n(x)$ can get arbitrarily big, hence Cauchy's criterion for uniform convergence doesn't hold, and the series doesn't uniformly converge. So the above theorem doesn't apply here. Is this theorem an "If and only if"? I couldn't find any mention for that. Is there another approach?
The series for $f$ does not converge uniformly on $(0,\infty)$. Both it and the series for $f'$ do converge uniformly on $[\delta, \infty)$ for any $\delta > 0$.
A more interesting series from this point of view is the Weierstrass function
$$ \sum_{n=1}^\infty a^{-n} \cos(7^{n} x) $$ where $(2 + 3 \pi)/14 < a < 1$. This series converges uniformly on $\mathbb R$ to a continuous function that turns out to be nowhere differentiable (but the latter fact is not at all obvious).
The problem in general is that given a series for $f(x)$, the convergence of the term-by-term derivative is is not necessary for $f'(x)$ to exist. Consider the series
$$ g(x) = \sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n} \sin(nx)$$ It can be shown that this converges uniformly to $x/2$ on $[-r,r]$ for any $r < \pi$, and that sum is of course differentiable. However, the term-by-term derivative series
$$ \sum_{n=1}^\infty (-1)^{n-1} \cos(nx) $$
diverges everywhere.