Consider $$\sum_{n=-\infty}^\infty c_n e^{ixn}$$ What's the difference between this series being absolutely convergent for $x = 0$ and just being absolutely convergent?
I mean, for $x = 0$, the series of interest is $\sum |c_n|$, but for all $x$, the series of interest is $\sum |c_n e^{ixn} |$ and the terms are less than $|c_n|$.
So if $\sum |c_n|$ converges, the series converges absolutely for all $x$, but this is already implied by absolute convergence for $x = 0$??
On
The series $\sum _{n=0}^{\infty }c_{n}e^{ixn}$ at $x=0$ is $\sum _{n=0}^{\infty }c_{n}$, which is not the same thing as $\sum _{n=0}^{\infty }\vert c_{n}\vert $ unless $c_{n}\geq 0$. So at $x=0$ you simply have a series of real (or complex) numbers which may or may not converge.
Now, it IS true to say that if the series $\sum _{n=0}^{\infty }\vert c_{n}e^{ixn}\vert $ converges then the original one does also and since $\vert c_{n}e^{ixn}\vert =\vert c_{n}\vert $ then using the fact that if a series converges absolutely, it converges, the result is this:
If $\sum _{n=0}^{\infty }\vert c_{n}\vert $ converges then $\sum _{n=0}^{\infty } c_{n}e^{ixn} $ does as well.
Testing for absolute convergence at any point $x$ will result in the same condition, namely whether $$\sum_{n=-\infty}^\infty |c_n|=0 .$$
This is because $|e^{ixn}|=1$ for all $x \in \mathbb{R}$.