Consider function $f$ which $2\pi$-pereodic and absolutely contionous on $[-\pi, \pi]$.
Prove, that Fourier coefficients in series expansion $f(x) = \frac{a_0}{2}+\sum\limits_{n=1}^\infty(a_n\cos nx+b_n\sin nx)$ are $a_n=o(1/n)$ and $b_n=o(1/n)$.
I tried calculating $a_n = \int\limits_{-\pi}^\pi f(x)\cos(nx)dx$ by parts, in order to get $\int\limits_{-\pi}^\pi nf(x)\sin(nx)dx$ and prove that it tends to $0$, but didn't achieved any good results. Also, I don't know what properties of absolute continuity are useful here.
The idea of utilizing IbP is correct, but you would want to apply it the other way around: For $n \neq 0$,
\begin{align*} a_n &= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos(nx) \, \mathrm{d}x \\ &= \underbrace{\frac{1}{\pi} \left[ f(x) \frac{\sin(nx)}{n} \right]_{-\pi}^{\pi}}_{=0} - \frac{1}{\pi n} \int_{-\pi}^{\pi} f'(x) \sin(nx) \, \mathrm{d}x \\ &= - \frac{1}{\pi n} \int_{-\pi}^{\pi} f'(x) \sin(nx) \, \mathrm{d}x. \end{align*}
Here, IbP is applicable because $f$ is absolutely continuous. Then by the Riemann–Lebesgue lemma, we have $\int_{-\pi}^{\pi} f'(x) \sin(nx) \, \mathrm{d}x = o(1)$ and hence $a_n = o(1/n)$. A similar argument also shows that $b_n = o(1/n)$.