= Solution
The <absolute Fourier convergence from a square-integrable derivative> uses more than the pointwise estimate $|\widehat f(n)|=O(1/|n|)$. Periodic <integration by parts> gives
$$
\widehat{f'}(n)=in\widehat f(n).
$$
By <Bessel's inequality>,
$$
\sum_{n\ne0}n^2|\widehat f(n)|^2
=\sum_{n\ne0}|\widehat{f'}(n)|^2
\leq\|f'\|_2^2.
$$
Now apply the <Cauchy-Schwarz inequality>:
$$
\boxed{\sum_{n\ne0}|\widehat f(n)|
\leq\left(\sum_{n\ne0}n^2|\widehat f(n)|^2\right)^{1/2}
\left(\sum_{n\ne0}\frac1{n^2}\right)^{1/2}
\leq\frac{\pi}{\sqrt3}\|f'\|_2.}
$$
Adding the finite constant coefficient proves absolute summability. Since a continuously differentiable periodic function has $f'\in L^2$, all hypotheses of part (iii) hold and its <Fourier series> converges uniformly.
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