Let $f\in C^1[-\pi,\pi]$ st $f(-\pi)=f(\pi)$ and define $$a_n=\int^{\pi}_{-\pi} f(t)\cos nt dt\,$$ for $n \in\Bbb{N}$ . Then does the sequence $\{na_n\}$ converges? And does the series $\sum^{\infty}_{n=1} n^2|a_n|^2$ converges as $n\to \infty$.
I know that $a_n \to 0$ as $n\to \infty$ and if the series converges the $\{na_n\}$ must converge to $0$.
On
Yes, the first question is answered by the celebrated Riemann-Lebesgue Lemma.
Since the Fourier coefficients of $f'$ are $na_n$, by applying the Bessel inequality to $f'$ you get $\sum_n n^2 a_n^2 \leq \frac{1}{\pi} \int_{-\pi}^{\pi}|f'|^2$.
By integration by parts, we have: $$ n\, a_n = \int_{-\pi}^{\pi} f(t)\, n\cos(nt)\, dt = -\int_{-\pi}^{\pi} f'(t)\sin(n t)\,dt \tag{1}$$ and since $f'\in C^0([-\pi,\pi])$, the RHS of $(1)$ converges to zero by the Riemann-Lebesgue lemma. $f'\in C^0([-\pi,\pi])$ implies $f'\in L^2([-\pi,\pi])$, hence $$ \sum_{n\geq 1} n^2 a_n^2 <+\infty $$ follows from Bessel's inequality.