Knowing $f \in C^r(\mathbb{R})$ $$c_n(\hat{f}) = \frac{1}{2\pi} \int_{0}^{2\pi} f(x)e^{-inx}dx, n \in \mathbb{Z}$$ I am trying to show that $$\mid c_n \mid \leqslant C_r (1 + {\mid n \mid}^{-r}) $$ I have been struggling the whole evening and I still can't see how the estimate depends on the smoothness $r$ and $n$. Since I just replaced $e^{-inx}$ by $1$ and I really don't know any other ways to deal with complex numbers. Thank you very much for your hint.
*Edit, I just found out that I am asked to show that $$\mid c_n \mid \leqslant C_r (1 + {\mid n \mid})^{-r} $$
Hint: In general $c_n = o(|n|^{-r})$ for $2\pi$-periodic $f \in C^r(\mathbb{R})$ and getting the exact form for your bound requires a little manipulation.
For $f \in C^1$ and $n \neq 0$, integration by parts yields
$$\begin{align}|c_n| &= \frac{1}{2\pi}\left|i\frac{f(2\pi)e^{-i2\pi n} - f(0)}{n} - \frac{i}{n}\int_0^{2\pi} f'(x)e^{-inx} \, dx\right| \\ &\leqslant \frac{|f(2\pi) - f(0)|}{2\pi}+ \frac{1}{2\pi|n|}\int_0^{2\pi} |f'(x)| \, dx \\ &= \frac{1}{2\pi}\left|\int_0^{2\pi}f'(x) \, dx \right|+ \frac{1}{2\pi|n|}\int_0^{2\pi} |f'(x)| \, dx \\ & \leqslant \frac{1}{2\pi}\int_0^{2\pi}|f'(x)| \, dx \left(1 + |n|^{-1} \right) \end{align}$$
For $f \in C^r$ use repeated integration by parts.