= Solution
Let $S_N(x)=\sum_{n=1}^N[a_n\cos(nx)+b_n\sin(nx)]$. Under the first coefficient bounds,
$$
\sup_{x\in\mathbb R}|S_M(x)-S_N(x)|\le2c\sum_{n>N}n^{-1-\varepsilon}\longrightarrow0.
$$
This proves the uniform Cauchy condition on the whole <real line> and hence <uniform convergence> to $f$. It is the <Weierstrass M-test> applied with the summable majorant $2cn^{-1-\varepsilon}$. Every partial sum is continuous. To see continuity of the limit explicitly, at a chosen $x_0$ first choose $N$ with small uniform error; then make $|S_N(x)-S_N(x_0)|$ small by continuity. The inequality
$$
|f(x)-f(x_0)|\le|f(x)-S_N(x)|+|S_N(x)-S_N(x_0)|+|S_N(x_0)-f(x_0)|
$$
proves continuity. Also $S_N(x+2\pi)=S_N(x)$ for every $N$, so taking limits gives \b[$f(x+2\pi)=f(x)$].
Under the stronger coefficient bounds, the <derivative> series has summable uniform majorant $2cn^{-1-\varepsilon}$. Thus $S_N'$ converges uniformly to a continuous function $g$ given by that series. The original series also converges uniformly. On any finite interval, the <fundamental theorem of calculus> gives $S_N(x)-S_N(0)=\int_0^x S_N'(t)dt$. <Uniform convergence> permits passage to the limit inside this finite <integral>, since the error is at most $|x|\|S_N'-g\|_\infty$. Therefore $f(x)-f(0)=\int_0^x g(t)dt$, and continuity of $g$ gives $f'=g$. This justifies <Termwise differentiation of a Fourier series> rather than assuming it:
$$
\boxed{f'(x)=-\sum_{n\ge1}na_n\sin(nx)+\sum_{n\ge1}nb_n\cos(nx).}
$$
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