Let $\hat f(n) = \frac{1}{2\pi} \int_{-\pi}^\pi f(t)e^{-int} \, \text{d}t$ be the Fourier coefficients of $f$.
Theorem. $\hat f(n) = 0$ implies $f(t) = 0$ when $f$ is continuous at $t$.
Proof. By linearity, this is equivalent to showing that $\hat f(n) = 0$ for all $n$ implies $f = 0$. Assume the contrary. Then $\int_{-\pi}^{\pi} q(t)f(t) \, \text{d}t$ should always be 0, where $q$ is any trigonometric polynomial, but $f(0) \neq 0$.
The overall idea is to isolate the behavior of $f$ at $0$ via a trigonometric polynomial.
$f$ is continuous at $0$, so we can choose a $\delta < \frac{\pi}{2}$ such that $f(t) \geq f(0)/2$ when $|t| < \delta$. Pick $\epsilon > 0$ small enough that $p(t) = \epsilon + \cos(t) < 1 – \epsilon/2$ when $|t| > \delta$. Finally, choose $\eta < \delta$ with $p(t) > 1 + \epsilon/2$ when $|t| < \eta$ and set $p_k(t) = p(t)^k$.
First, $\int_{|t| \leq \eta} p_k(t)f(t)\, \text{d}t > (1 + \epsilon/2)^k f(0)/2 \to \infty$.
Second, $\int_{\eta < |t| \leq \delta} p_k(t)f(t)\, \text{d}t \geq (1 – \epsilon/2)^k f(0)/2 \geq 0$.
Third, $\left| \int_{\delta < |t| \leq \pi} p_k(t)f(t)\, \text{d}t \right| \leq (1 – \epsilon/2)^kM$ where $M$ is some bound on $f$. This quantity shrinks to 0.
Combining all three, we can see that $\int_{-\pi}^\pi p_k(t)f(t)\, \text{d}t \to \infty$ as $n$ gets large. By assumption, however, it should be zero. Contradiction. $\blacksquare$
Definition. Let $S_N(t) = \sum_{|n| \leq N}\hat f(n)e^{int}$ be the partial sums of the Fourier series of $f$.
Theorem. If $f$ is continuous and $\sum |\hat f(n)| < \infty$, then $S_N \to f$ uniformly.
Proof. As in the Weierstrass M-test to show uniform convergence (remember that $|e^{int}| = 1$) to some function $g$. Also note that the uniform limit of continuous functions is continuous, hence $g$ is continuous. Also, limiting sums commute with integrals under uniform convergence, implying
$$
\begin{eqnarray}
\hat g(n) &=& \frac{1}{2\pi} \int_{-\pi}^\pi g(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \int_{-\pi}^\pi \lim_N S_N(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \int_{-\pi}^\pi S_N(t)e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \int_{-\pi}^\pi \sum_{|m| \leq N} \hat f(m)e^{imt}e^{-int} \, \text{d}t \\
&=&\frac{1}{2\pi} \lim_N \sum_{|m| \leq N}\hat f(m) \int_{-\pi}^\pi e^{i(m-n)t}\, \text{d}t \\
&=& \hat f(n).
\end{eqnarray}
$$
Ergo $g = f$, since their Fourier coefficients are the same and both are continuous everywhere.
Theorem. If $f \in C^1(\Bbb T)$, then $S_n \to f$ uniformly.
Proof. $\frac{1}{2\pi} \int_{-\pi}^\pi f(t)e^{-int} dt = \frac{1}{in}\frac{1}{2\pi}\int_{-\pi}^\pi f'(t)e^{-int} dt$ through integration by parts (this is where continuity of $f’$ is needed), so that $\hat{f}(n) = \frac{1}{in} \widehat{f’}(n)$. Furthermore, $\sum |\widehat{f’}(n)|^2 =\frac{1}{2\pi} \int_{-\pi}^\pi |f'(t)|^2 dt < \infty$ by Parseval’s identity. Therefore $\sum |\hat f(n)| \leq \left( \sum \frac{1}{n^2} \right) \left( | \widehat{f’}(n)|^2 \right) < \infty$ by Cauchy-Schwarz. $\blacksquare$
I think $|c_n| \leq 1/n^{k+1+\epsilon}$ for $\epsilon > 0$ implies $f \in C_k([-\pi, \pi])$.