If $f$ is periodic and has derivative that is sectionally continuous(piecewise continuous) then $\lim_{|n|\rightarrow\infty} n\hat{f}(n)=0$
Just need someone to explain to me in layman's terms why I need the stated hypothesis for the problem. I'm confused as to why we need periodicity of $2\pi$ and a sectionally continuous derivative if we are just showing fourier coefficients converge to zero, will I need the fact that it will be bounded above to show convergence to $0$? I'm assuming the reason that $|n|\rightarrow\infty$ is to say that there is no fourier transform representation, am I understanding the preface for developing a proof? If you dislike my question please justify, no proof does not imply bad question
Update with proof attempt: $2\pi\hat{f}(n)=\int_0^{2\pi}f(\theta)e^{-in\theta}d\theta=\frac{1}{in}\int_0^{2\pi}f'(\theta)e^{-in\theta}d\theta$, where we obtained the second equality by integration by parts and where the first term of the integration, which I did not post, goes to zero since we have periodicity. So $2\pi|n||\hat{f}(n)|\leq|\int_0^{2\pi}f'(\theta)e^{-in\theta}d\theta|\leq\int_0^{2\pi}|f'(\theta)|d\theta\leq C$ where C is independent of $n$ and we are done.
I assume you meant that $f$ is $2\pi$-periodic and the primitive of a piecewise continuous $2\pi$-periodic function, denoted $f'$.
You only proved that $|n \hat{f}(n)|$ is bounded, not that it $\to 0$.
For this, show that as $n\to \infty$ $$\frac1{2\pi}\int_0^{2\pi} |f'(x)-f'(x+\pi/n)|dx\to 0$$ Then use
$$2 \widehat{f'}(n)=\frac1{2\pi}\int_0^{2\pi} (f'(x)-f'(x+\pi/n)) e^{-inx}dx$$