Given the series $\sum_{k \in \mathbb{Z}}c_k e^{ikx}$. Show that if the coefficients $c_k$ satisfy: There exists a constant L>0 such that $\forall k \in \mathbb{Z}$ $\exists m \in \mathbb{N} : |c_k|\leq Lk^m$ Then the series converges in $\mathcal{D'}(\mathbb{R})$
First I looked up the Definition of "converges in $\mathcal{D'}(\mathbb{R})$":
For $A_n \in \mathcal{D'}(\mathbb{R})$ and $A \in \mathcal{D'}(\mathbb{R})$ the sequence $A_n \rightarrow A$ in $\mathcal{D'}(\mathbb{R})$ if $\langle A_n,f \rangle \rightarrow \langle A,f \rangle$ for every testfunction $f \in \mathcal{D'}(\mathbb{R})$.
The second convergence is the usual convergence in $\mathbb{R}$.
First I rewrote the series so It looks "nicer":
$\sum_{k \in \mathbb{Z}}c_k e^{ikx}=lim_{n \rightarrow \infty} (\sum_{k=0}^{n}c_k e^{ikx} +\sum_{k=1}^{n}c_{-k} e^{-ikx} )$ As I don't know to which function the series converges I thought that I would try to show that it is a Cauchy series. This means I need to show $|\langle A_p,f \rangle - \langle A_q,f \rangle| \leq \epsilon$ where $p,q \in \mathbb{N}$.
$|\langle A_p,f \rangle - \langle A_q,f \rangle|= |\int_{-\infty}^{\infty}f(x) (\sum_{k=q}^{p}c_k e^{ikx} +\sum_{k=q}^{p}c_{-k} e^{-ikx}) dx| \leq \int_{-\infty}^{\infty}|f(x)| (|\sum_{k=q}^{p}c_k e^{ikx}| +|\sum_{k=q}^{p}c_{-k} e^{-ikx}|) dx \leq \int_{-\infty}^{\infty}|f(x)| (\sum_{k=q}^{p}Lk^m +\sum_{k=q}^{p}Lk^m ) dx= 2L \int_{-\infty}^{\infty}|f(x)| \sum_{k=q}^{p}k^m )$
I don't really know how to continue. I would appreciate it if someone could help me.
Edit:
Using the hint I got from Alex Ortiz
Let $f \in \mathcal{D}(\mathbb{R})$
$\langle e^{ikx},f(x) \rangle=\int_{-\infty}^{\infty}e^{ikx}f(x)dx=-\int_{-\infty}^{\infty}\frac{e^{ikx}}{ik}f(x)dx$
Let M<N
$lim_{M,N \rightarrow \infty} \sum_{M<|k|<N}c_k \langle e^{ikx},f(x) \rangle \leq lim_{M,N \rightarrow \infty} \sum_{M<|k|<N}|c_k| \langle e^{ikx},f(x) \rangle \leq lim_{M,N \rightarrow \infty} \sum_{M<|k|<N}L|k|^m \langle e^{ikx},f(x) \rangle = lim_{M,N \rightarrow \infty} \sum_{M<|k|<N}L|k|^m (-1)^{m+1} \int_{-\infty}^{\infty}\frac{e^{ikx}}{(i)^{m+1} (k)^{m+1}}f(x)dx=im_{M,N \rightarrow \infty} \sum_{M<|k|<N}\frac{L}{k} (-1)^{m+1} \int_{-\infty}^{\infty}\frac{e^{ikx}}{(i)^{m+1} }f(x)dx$
More in general, I can partially integrate as often as I like. This way I can guarantee that for every $\epsilon$ and all m,n the absolute value of the partial sum is still less then $\epsilon$
Did you copy the assumption about the coefficients down correctly? The order of the quantifiers matters. $$\exists L\ \forall k\ \exists m:|c_k|\le L|k|^m$$ is the order you have written down, but I suspect is not what was intended since you can choose the sequence $\{c_k\}$ in such a manner that the result doesn't hold ($c_k = |k|^{|k|}$ should suffice as a counterexample). I believe the assumption should be $$\exists L\ \exists m\ \forall k:|c_k|\le L|k|^m.$$ In other words, the sequence of coefficients has at most polynomial growth.
You are correct about the definition of convergence in $\mathcal D'$. In this case, it amounts to proving that for any fixed test function $f$, the sequence of real numbers $\{\langle \sum_{|k|<N}c_ke^{ikx},f(x)\rangle\}_{N=1}^\infty$ is Cauchy, i.e., $$ \lim_{M,N\to\infty}\sum_{M<|k|<N}c_k\langle e^{ikx},f(x)\rangle=0. $$ Here is a sketch of how to proceed. When you evaluate, $\langle e^{ikx},f(x)\rangle=\int_{-\infty}^\infty e^{ikx}f(x)\,dx$.
Fact. Since $f$ is a fixed test function, there is some $A>0$ so that $f(x)=0$ for all $|x| > A$.
Using the fact, try integrating $\int_{-\infty}^\infty e^{ikx}f(x)\,dx$ by parts and see what you get.