Let $f$ be a smooth periodic function $C^{\infty}([0,2\pi])$. Let $$ I=\int_{0}^{2\pi}f(x)dx.$$ The trapezoidal rule is given by $$ I_{N}=\frac{2\pi}{N}\sum_{k=0}^{N-1}f(x_{k}).$$ Then,
$$ I_{N}=I+2\pi\sum_{r \neq 0}\hat{f_{rN}},$$ where $$ \hat{f_{k}}=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{ikx}dx.$$ We know that for any integer $p >0$, and for any integer $k \neq 0$, that there exists a constant $C_{p}$ such that $$ |\hat{f_{k}}|\le \frac{C_{p}}{|k|^{p}}.$$
Prove that the error in the Trapezoidal rule for a smooth period function decays faster than any polynomial as $N \to \infty$.
Any guidance on how to prove this?