Let's say I have a piecewise continuous function which has the fourier series $\sum_n\ c_{n}e^{inx}$ and I assume that $ \sum_n\ n|c_{n}|$ converges, then I know the following holds:

The fourier series of the derivatives $\sum_n\ inc_{n}e^{inx}$ is absolutely convergent and thus from the Weierstrass M-test I know $\sum_n\ inc_{n}e^{inx}$ is uniformly convergent.

Thus, $ f' (x)=\sum_n\ inc_{n}e^{inx}$ is a continuous $2 \pi $-periodic function.

We can say then that $ f(x)=\sum_n\ c_{n}e^{inx}$ is a continuously differentiable $2 \pi $-periodic function.

My question is this:

Does the converse also holds?

To be more precise:

If $ f(x)=\sum_n\ c_{n}e^{inx}$ is a continuously differentiable $2 \pi $-periodic function, does it follow that $\sum_n\ n|c_{n}|$ converges?

My thought on the matter are such:

I think the converse does not hold, since we've used the Weierstrass M-test which is kind of a one-way ticket.

I attempted at constructing a counter-example. Since I want $\sum_n\ n|c_{n}|$ to diverge, it suffices to find $c_{n}$ such that $c_{n} \sim \frac{1}{ n^{2} } $.

Thus, if I look at $ f' (x)=\sum_n\ inc_{n}e^{inx}$, it is clear that I need a continuous $2 \pi $-periodic function such that its fourier coefficients decay rate is $\frac{1}{ n } $.

Unfortunately, every continuous function I tried had a fourier coefficients decay rate of $\frac{1}{ {n^{2} } } $. The ones that I know of that have a decay rate of $ \frac{1}{ n } $ are discontinuous functions.

Your thoughts on the matter are greatly appreciated!

Here is a counterexample. This web page https://en.wikipedia.org/wiki/Wiener_process#Wiener_representation gives the randomly created function $$ W(t) = \zeta_0 t + \frac{\sqrt 2}{\pi} \sum_{n=1}^\infty \frac {\zeta_n \sin(n \pi t)}n $$ where $(\zeta_n)$ is a sequence of independent random variables distributed Gaussian mean 0, variance 1. This page says this is one way to create Brownian motion on $[0,1]$. Thus $$f(t) = W(t) - W(1) t = \frac{\sqrt 2}{\pi} \sum_{n=1}^\infty \frac {\zeta_n \sin(n \pi t)}n $$ is continuous with probability 1. But one can show that $$ \sum_{n=1}^\infty \frac{|\zeta_n|}{n} $$ diverges with probability 1 (hint: first show that $ \sum_{n=1}^\infty \frac{|\zeta_n|-\mathbb E|\zeta_n|}{n} $ converges using Kolmogorov's three series theorem https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem).

Further notes: Another place to look for counterexamples is with random functions of the form $$ f(x) = \sum_{n=-\infty}^\infty r_n a_n e^{i\pi n x} $$ where $r_n$ are identically distributed random variables taking values plus or minus one with probability $1/2$. If you look in the book Random Fourier Series with Applications to Harmonic Analysis by Michael B. Marcus & Gilles Pisier, or in the book Some Random Series of Functions, 2nd Edition by Jean-Pierre Kahane, you will find conditions on the sequence $a_n$ such that the random function is continuous almost surely. Then out of those sequences, it should be relatively straightforward to find an example that is not absolutely summing.

Also, look at https://en.wikipedia.org/wiki/Wiener_algebra, but that only tells you the name of the space of functions whose coefficients are absolutely summing - it doesn't provide examples of continuous functions not in the Wiener algebra.

Suppose $h$ is a periodic continuous function $[0,2\pi]$ for which $\sum_n |c_n| =\infty$. Define $$ f(x) = \int_0^x \left[h(t)-\frac{1}{2\pi}\int_0^{2\pi}h(u)du\right]dt. $$ Then $f$ is continuously differentiable and periodic on $[0,2\pi]$ with Fourier coefficients $c_n(f)$ for which \begin{align} inc_n(f) & = -\frac{1}{2\pi}\int_{0}^{2\pi}f(y)(-ine^{-iny})dy \\ & = -\frac{1}{2\pi}\int_{0}^{2\pi}f(y)\frac{d}{dy}e^{-iny}dy \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}h(y)e^{-iny}dy = c_n(h),\;\;\; n\ne 0. \end{align} So your problem reduces to considering the Fourier series of a periodic continuous function $h$. It is known that there is a periodic continuous function $h$ such that the Fourier series for $h$ diverges at $0$. Such an $h$ provides the counter example that you want because $\sum_n|c_n(h)| =\infty$ must occur.