Suppose that $2\pi$ periodic function $f$ has continuous second derivatives. Then the series $\sum_{n=-\infty }^{\infty } a_n e^{-inx}$ converges for each fixed $x\in [-\pi ,\pi ]$ where $a_n=\int_{-\pi }^{\pi }f(x)e^{-inx}dx$
My attempt: I used the integration by parts twice.
$a_n=\int_{-\pi }^{\pi }f(x)e^{-inx}dx=\int_{-\pi }^{\pi }f(x)\frac{\mathrm{d} \frac{e^{-inx}}{-in}}{\mathrm{d} x}dx= \Big[e^{-inx}f(x)\Big]_0^{\infty}-\int_{-\pi }^{\pi }e^{-inx}f'(x)dx=-\int_{-\pi }^{\pi }e^{-inx}f'(x)dx =-\int_{-\pi }^{\pi }f'(x)\frac{\mathrm{d} \frac{e^{-inx}}{-in}}{\mathrm{d} x}dx =\int_{-\pi }^{\pi }e^{-inx}f''(x)dx $
But,I don't know how to show the boundedness and do the comparison test from here.
Hint
There is a missing $n$ in your integration by parts: $$\int_{-\pi}^\pi f(x)e^{-inx} dx=-\color{red}{\frac{1}{-in}} \int_{-\pi}^\pi f'(x)e^{-inx} dx$$ so when you do two integrations by parts: $$\int_{-\pi}^\pi f(x)e^{-inx} dx=\color{red}{\frac{-1}{n^2}} \int_{-\pi}^\pi f''(x)e^{-inx} dx$$ and: $$\sum_n \frac{1}{n^2} < + \infty$$