Consider $f:[-\pi, \pi] \to \mathbb{C}$ is analytic (infinite differentiable) and periodic. Define $a_n:= \frac{1}{2 \pi}\int_{-\pi}^\pi f(x) e^{-inx}dx$ (the Fourier coefficient of $f$). Show that the sequence $\{n a_n\} \to 0$ as $n \to \infty$.
Here is my thinking: $$\begin{array}{l} |n{a_n} - 0| = \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi {nf} (x){e^{ - inx}}dx} \right| \le \frac{1}{{2\pi }}\int_{ - \pi }^\pi {\left| {nf(x){e^{ - inx}}} \right|} dx\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{} \end{array} \le \frac{1}{{2\pi }}\left\| f \right\|\int_{ - \pi }^\pi {\left| {n{e^{ - inx}}} \right|} dx \end{array}$$ Then I stuck to infer it further (since I can't say the integral above will converges eventually.)
Do not use the triangle inequality to bring the absolute value inside the integral — it will not help, as $\lvert e^{-inx}\rvert =1$. Instead, use integration by parts. $$ \int_{ - \pi }^\pi nf (x)e^{- inx} dx = \left[i f(x)e^{- inx}\right]^\pi_{-\pi} -i \int_{[-\pi,\pi]} f^\prime(x) e^{- inx}dx $$ For the first term, leverage the periodicity of $f$. For the second, use for instance the Riemann-Lebesgue Lemma.