This is only part of a larger question.
I've already shown that if $f$ is $2\pi$ periodic, then $$c_n = \frac{1}{4\pi}\int_{-\pi}^\pi\left(f(x)-f(x+\pi/n)\right)e^{-inx}dx.$$
I'm asked to then show that if $f$ is also lipschitz continuous then there is some constant $M$ such that $|c_n|\leq \frac{M}{|n|}$.
Using the lipschitz property, we then have $$\frac{1}{4\pi}\int_{-\pi}^\pi |f(x)-f(x+\pi/n)|e^{-inx}dx\leq \frac{1}{4\pi}\int_{-\pi}^\pi M\left\vert \frac{\pi}{n}\right\vert e^{-inx}dx = \frac{M}{4|n|}\int_{-\pi}^\pi e^{-inx}dx.$$
However, $\int_{-\pi}^\pi e^{-inx}dx = 0$, so unless the fourier coefficients are all $0$, I'm pretty sure that I'm missing something.
Any thoughts would be really appreciated.
Thanks in advance!
Your inequalities don't even make sense since there is no ordering for complex numbers. To correct your argumnet all you have to do is to insert absolute value sign for $e^{-inx}$.