$\text { The series } \sum _ { n = 1 } ^ { \infty } \sec ( n \pi ) \text { converges. }$

Question

Quizplus · Accepted Answer

The terms of the series alternate between $	ext{sec}(2n\pi)$ and $	ext{sec}((2n+1)\pi)$ for $n\geq 1$. Since $	ext{sec}(2n\pi) = 1$ for all $n$, it suffices to consider the terms $	ext{sec}((2n+1)\pi)$. Note that $	ext{sec}((2n+1)\pi) = (-1)^n$ for all $n$, so the series is equivalent to the divergent series $\sum_{n=1}^\infty (-1)^n$. Therefore, the series $\sum_{n=1}^\infty 	ext{ sec}(n\pi)$ diverges.

The series ∑n=1∞sec⁡(nπ) converges. \text { The series } \sum _ { n = 1 } ^ { \infty } \sec ( n \pi ) \text { converges. } The series ∑n=1∞​sec(nπ) converges.

$\text { The series } \sum _ { n = 1 } ^ { \infty } \sec ( n \pi ) \text { converges. }$