I have seen some similar questions around, but they use a slightly different assumption than mine. Furthermore, I am trying to figure out the details of the proof, and I got what I believe is a mistake.
I am trying to prove the following.
Define the function
$$f(x) = \sum_{k \in \mathbb{Z}} c_k e^{2 \pi i kx}$$
where $x \in \mathbb{R}$ and assume that $\forall k, \ |c_k| \leq ae^{-bk^2}$ for positive real numbers $a,b$.
Then, $f$ is continuous, and infinitely differentiable. Also, find an expression for $f'(x)$.
For continuity I just used well-known inequalities. Assume $x \to x_0$:
$$|f(x)-f(x_0)| = \left| \sum_k c_k \left( e^{2 \pi i k x} - e^{2 \pi i k x_0} \right) \right| \leq \sum_k \left| c_k \left( e^{2 \pi i k x} - e^{2 \pi i k x_0} \right) \right| = \sum_k |c_k| \left| e^{2 \pi i k x} - e^{2 \pi i k x_0} \right| \leq \sum_k ae^{-bk^2} \left| e^{2 \pi i k x} - e^{2 \pi i k x_0} \right| \to0 \ \ \ \ \ \ \text{ as } \ \ x \to x_0$$
Now, let's just try to prove that $f$ is once-differentiable.
If I try to do it by the definition of limits, I get
$$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\sum_k c_k\left( e^{2 \pi i k (x+h)} - e^{2 \pi i k x} \right)}{h} $$
and I can now just put the limit "through" the sum which is basically just obtaining the limit that gives the derivative of the exponential function, which is known to be $\frac{d}{dx}e^{rx} = re^{rx}$
So this gives $$\sum_k c_k \lim_{h \to 0}\frac{e^{2 \pi i k (x+h)} - e^{2 \pi i k x}}{h} = \sum_k c_k(2\pi i k)e^{2 \pi i kx}$$
Similarly, for proving that it is infinitely differentiable I can do that by induction and use the same procedure as above to show that the $n$th derivative is just
$$\sum_k c_k (2 \pi i k)^n e^{2 \pi i k x}$$
My problem is that I never used the assumption that the coefficients are rapidly decaying for the differentiability proofs, so I am thinking I got something wrong, but I don't know what.