Show that the series $\sum_{n=0}^{\infty}\dfrac{\cos(nx)}{\cos^{n}(x)}$ is divergent

Question

In this thread, I use the "C+iS" summation method for trigonometric series to sum it:

lab bhattacharjee pointed out that:

$\left|\dfrac{e^{ix}}{\cos(x)}\right|=|\sec(x)|\geq1$

What I don't understand is why $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, I thought it is $e^{ix}\sec(x)$. By looking at the graph of $\sec(x)$, it is clear that $\sec(x)\geq1 $ on the interval $(0,\pi/2)$ and $(0,5\pi/2)$ and so on. $\sec(x)\leq1$ on the interval $(\pi, 3\pi/2)$ and $(3\pi/2, \pi)$. So I get the part if $\dfrac{e^{ix}}{\cos(x)}=\sec(x)$, then the series diverge.

Also, how can I show that the series $\left(\dfrac{e^{-ix}}{\cos(x)}\right)^n$ is a divergent series?

Answer 1

Note that $$\left|\dfrac{e^{ix}}{\cos x}\right| = \left|\frac{\cos x + i\sin x}{\cos x}\right| = \left| 1 + i\tan x\right| = \sqrt{1+\tan^2 x} = \left|\sec x\right|$$

Answer 2

$$\lim_{n\to\infty}\frac{\cos (n x)}{\cos ^n(x)}$$ fails to exist, therefore the general term does not tend to zero, necessary condition for the convergence. Thus the series diverges.