Find out the value of x for which the given series convergent?

Question

Given series

$\sum_{n=1}^{\infty} 2^n (\tan x)^{n^2}$

Find out the real numbers for which the following integral is convergent?

For solution i take $a_n=2^n (\tan x)^{n^2}$

and apply Cauchy root test $(a_n)^{1/n}= 2(\tan x)^{n}$

Now for convergence $\lim_{ n\to \infty}$ $|a_n| \lt 1$

Thus $\lim_{n\to \infty}(\tan x)^{n}\lt \frac{1}{2}$

Now What can we say about $x$?

Answer 1

Following up from your steps

$\lim_{n \rightarrow \infty} |\tan x|^n < \frac{1}{2}$
$\Rightarrow |\tan x| < 1$
$\Rightarrow x \epsilon (n\pi -\frac{\pi}{4} , n\pi + \frac{\pi}{4})$