For a test function in Schwartz space $\mathscr{S}(\mathbb{R})$, let me define the Fourier transform as
$$
\widehat{f}(\xi):=\int_{\mathbb{R}}e^{-i\xi x}f(x)\ dx\ .
$$
For a distribution $T$ in the space of temperate distributions $\mathscr{S}'(\mathbb{R})$, the Fourier transform $\widehat{T}$ will then be given by
$$
\langle \widehat{T},f\rangle:=\langle T,\widehat{f}\rangle\ .
$$
It is easy to see that, as temperate distributions (or generalized functions of $x$)
$$
\sum^{n}_{k=-n} c_k\ e^{2i\pi kx}=\widehat{A_n}(x)
$$
where $A_n$ is the temperate generalized function of $\xi$ given by
$$
A_n(\xi):=\sum^{n}_{k=-n} c_k \ \delta(\xi +2\pi\ k)\ .
$$
Namely, for every $f\in\mathscr{S}(\mathbb{R})$,
$$
\langle A_n,f\rangle:=\sum^{n}_{k=-n} c_k \ f(-2\pi k)\ .
$$
Now let me go ahead and define
$$
\langle A_{\infty},f\rangle:=\sum_{k\in\mathbb{Z}} c_k \ f(-2\pi k)\ .
$$
I will first show that $A_{\infty}$ is well defined and belongs to $\mathscr{S}'(\mathbb{R})$.
I will use the notation $\langle x\rangle=\sqrt{1+x^2}$ for the so-called Japanese bracket.
By Cauchy-Schwarz we have $1+|x|\le \sqrt{2}\ \langle x\rangle$. We also have the trivial inequality $1+|\xi|\le 1+|-2\pi\xi|$.
For nonnegative integers $p,q$ let me define the seminorms
$$
||f||_{p,q}=\sup_{x\in\mathbb{R}}\ \langle x\rangle^q\ |f^{(p)}(x)| \ .
$$
These define the topology of $\mathscr{S}(\mathbb{R})$.
Now we have
$$
|\langle A_{\infty},f\rangle|\le \sum_{k\in\mathbb{Z}} |c_k| \ |f(-2\pi k)|
$$
$$
\le \sum_{k\in\mathbb{Z}} C (1+|k|)^{-2} (1+|k|)^{m+2}\ |f(-2\pi k)|
$$
$$
\le \sum_{k\in\mathbb{Z}} C (1+|k|)^{-2} 2^{\frac{m+2}{2}} \langle -2\pi k\rangle^{m+2}\ |f(-2\pi k)|
$$
$$
\le C\times\left(\sum_{k\in\mathbb{Z}} (1+|k|)^{-2} \right)\times 2^{\frac{m+2}{2}}\times ||f||_{0,m+2}\ .
$$
These bounds show the absolute convergence of the series defining $\langle A_{\infty},f\rangle$.
The latter is clearly linear in $f$. Most importantly, we bounded $|\langle A_{\infty},f\rangle|$ by a constant times the seminorm $||f||_{0,m+2}$ which proves continuity in $f\in\mathscr{S}(\mathbb{R})$. So $A_{\infty}\in\mathscr{S}'(\mathbb{R})$.
The correct topology to use systematically on $\mathscr{S}'(\mathbb{R})$ is the strong topology and not the weak-$\ast$ topology. A subset $B\in\mathscr{S}(\mathbb{R})$ is called bounded if and only if for all $p,q$,
$$
\sup_{f\in B}||f||_{p,q}\ <\infty\ .
$$
The strong topology on $\mathscr{S}'(\mathbb{R})$ is the one defined by the seminorms
$$
||T||_{B}=\sup_{f\in B} |\langle T,f\rangle|
$$
where $B$ ranges over bounded subsets of $\mathscr{S}(\mathbb{R})$.
Let $B$ be such a bounded set and let $f\in B$. Then by the repeating the previous estimates, now with the extra condition $|k|>n$ on the summation index,
$$
|\langle A_n,f\rangle-\langle A_{\infty},f\rangle|\le
C\times\left(\sum_{k\in\mathbb{Z},|k|>n} (1+|k|)^{-2} \right)\times 2^{\frac{m+2}{2}}\times ||f||_{0,m+2}\ .
$$
Therefore
$$
||A_n-A_{\infty}||_B\le
C\times\left(\sum_{k\in\mathbb{Z},|k|>n} (1+|k|)^{-2} \right)\times 2^{\frac{m+2}{2}}\times
\sup_{f\in B} ||f||_{0,m+2}
$$
and consequently $||A_n-A_{\infty}||_B\rightarrow 0$ when $n\rightarrow\infty$ because the series
$$
\sum_{k\in\mathbb{Z}} (1+|k|)^{-2}
$$
converges. The above argument proves that $A_n$ converges to $A_{\infty}$ in $\mathscr{S}'(\mathbb{R})$.
To finish, invoke the continuity of the Fourier transform $\mathscr{S}'(\mathbb{R})\rightarrow \mathscr{S}'(\mathbb{R})$
and the continuity of the canonical injection $\mathscr{S}'(\mathbb{R})\rightarrow \mathscr{D}'(\mathbb{R})$. Again, the spaces $\mathscr{S}'(\mathbb{R})$ and $\mathscr{D}'(\mathbb{R})$ carry the strong topology. In the case of $\mathscr{D}'(\mathbb{R})$, one first needs to understand the topology of $\mathscr{D}(\mathbb{R})$ which is explained in
Doubt in understanding Space $D(\Omega)$
Then one needs to understand what are the bounded subsets in $\mathscr{D}(\mathbb{R})$. These are sets of test functions whose supports are contained in a common compact set and whose derivatives are bounded uniformly (in the function chosen in the set, not in the order of derivation).
