Prove that $\sum_{n=0}^{\infty}e^{in\theta}$ is bounded

Question

For my homework class, we need to prove that a certain series converges, for which it is useful to use that this series is bounded ($\theta \in (0,2\pi)$):

$$\sum_{n=0}^{\infty}e^{in\theta}$$

I think that for all $m$

$$\left|\sum_{n=0}^{m}e^{in\theta}\right|\leq \frac{2}{\theta}+1$$

But I do not see how I can prove this.

edit for removing the max function, increasing simplicity

Answer 1

Expanding on DavidH's hint. Since $\theta \ne 0, 2\pi$, $e^{i\theta} \ne 1$. Therefore we use the geometric series $$ \sum_{n=0}^N e^{in\theta} = \frac{1 - e^{i(N+1)\theta}}{1 - e^{i\theta}} $$ So $$ \left| \sum_{n=0}^N e^{in\theta} \right| = \left| \frac{1 - e^{i(N+1)\theta}}{1 - e^{i\theta}} \right| \le \frac{2}{|1 - e^{i \theta}|}\in \mathbb{R}. $$