Showing that we can exchange infinite sum with limit I am trying to prove that the $n$th derivative of $f$ exists for 
$$f(x) = \sum_{k \in \mathbb{Z}}c_ke^{2 \pi ikx}$$
where $|c_k| \leq ae^{-bk^2}$ for each $k$ and some values $a,b > 0$
I have problems from the first derivative. What I tried to do was
$$\lim_{h \to 0} \sum_{k \in \mathbb{Z}} \frac{e^{2 \pi i k (x+h)} - e^{2 \pi i kx}}{h} = \lim_{h \to \infty} \int_\mathbb{Z} c_k \frac{e^{2 \pi i k (x+1/h)} - e^{2 \pi i kx}}{1/h} \mathrm{d}\delta(k) $$
where $\delta$ is the counting measure.
My problem is that, I need to be able to interchange the sum (or integral) and the limit. I planned on using the Dominated Convergence Theorem, but I have the following problem
$$|c_k| \left| \frac{e^{2 \pi i k (x+1/h)} - e^{2 \pi i kx}}{1/h} \right| \leq ae^{-bk^2} \cdot2\cdot\frac{1}{|1/h|}$$
But I can't get rid of the $h$ to have an integrable function $g$ such that $g(x) \geq \left| c_k \frac{e^{2\pi i k(x+1/h)}-e^{2 \pi i k x}}{1/h} \right|$ and that does not depend on $h$. I know I am supposed to use the assumption involving $ae^{-bk^2}$ but I can't seem to make it work.
Thank you!

Do not change $h$ to $\frac  1 h$. Just use the inequality $|e^{it}-1|\ \leq |t|$ ($t$ real ) to get a dominating sequence of the type $C |kc_k|$ where $C$ is a constant. 
