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I suggest this problem to you.

Let $\theta \in \mathbb{R}$. For all nonnegative integer $n$ let $s_{n}=1+\exp(i\theta)+\exp(2i\theta)+\ldots + \exp(in\theta)$. Determine a necessary and sufficient condition over $\theta$ for the sequence ${(s_{n})}_{n \in \mathbb{N}}$ to be bounded.

Terminology : The complexe sequence $(s_{n})_{n \in \mathbb{N}}$ is said to be bounded iff $$\exists M \in \mathbb{R}^{+}\, \forall n \in \mathbb{N} \left | z_{n}\right | \leq M.$$

Any idea ?

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1 Answer 1

Let $s_n(\theta)=\sum_{k=0}^ne^{ik\theta}$. If $\theta\in 2\pi \mathbb Z$ then $e^{i\theta}=1$ and $s_n(\theta)=n$, so $\{s_n\}$ is not bounded. If $\theta\notin 2\pi\mathbb Z$, then $$s_n(\theta)=\frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}=\frac{e^{i\frac{n+1}2\theta}}{e^{i\frac{\theta}2}}\frac{e^{i\frac{n+1}2\theta}-e^{-i\frac{n+1}2\theta}}{e^{i\frac{\theta}2}-e^{i\frac{\theta}2}}=e^{i\frac{n\theta}2}\frac{\sin\left(\frac{n+1}2\theta\right)}{\sin\frac{\theta}2},$$ so $$|s_n(\theta)|\leq \frac 1{\left|\sin\frac{\theta}2\right|}.$$ Finally, the sequence $\{s_n(\theta)\}$ is bounded if and only if $\theta\notin 2\pi\mathbb Z$.

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