Convergence of $\left(e^{i n \theta}\right)_{n \in \mathbb{N}}$

Show that the sequence $\left(e^{i n \theta}\right)_n$ converges if and only if $\theta \equiv 0(\bmod 2 \pi)$ .

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Solution by SequoiaFR

Discussions2 useful votes

On note $l$ la limite de $(e^{in\theta})_n$ . Ainsi $(e^{i(n+1)\theta})_n$ est une sous-suite de $(e^{in\theta})_n$ donc converge aussi vers $l$ . Cependant on a aussi que $e^{i(n+1)\theta}=e^{i\theta}e^{in\theta}$ qui va alors tendre vers $e^{i\theta}l$ à l’infini.

Or, par continuité de $z\longmapsto\vert z\vert$ , on a que $\vert l\vert=1$ donc $l\neq0$ et l’unicité de la limite nous donne donc que $e^{i\theta}=1$ , soit que $\theta=0\,[2\pi]$ .

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