Find the interval and radius of convergence of the series 
Determine the radius and interval of convergence of $$\sum_{n=1}^{\infty} \frac{(-1)^nx^n}{4^n\ln(n)}$$

We apply the ratio test, thus
$$\lim_{n \to \infty} \bigg| \frac{\frac{x^{n+1}}{4^{n+1}\ln(n+1)}}{\frac{x^{n}}{4^{n}\ln(n)}} \bigg| = \lim_{n\to \infty} \bigg|\frac{x\ln(n)}{\ln(n+1)4} \bigg| = |x| \lim_{n \to \infty} 1/4 < 1$$
So $|x| < 4$?
 A: The ratio test gives the limit $L=\frac{|x|}{4}$.
so, $|x|<4 \implies L<1 \implies $ the series converges.
and
$|x|>4 \implies L>1 \implies $ the series diverges.
thus the radius of convergence is $R=4$.

at $x=4$



*

*$\lim_{n\to+\infty}\frac{1}{\ln(n)}=0$

*$n\mapsto \frac{1}{\ln(n)}$ is decreasing, so by alternate series criterion, it converges.

at $x=-4$.

$\lim_{n\to +\infty}\sqrt{n}\frac{1}{\ln(n)}=+\infty$
$\implies \sqrt{n}\frac{1}{\ln(n)}>1 $ for large enough $n$
$\implies \frac{1}{\sqrt{n}}<\frac{1}{\ln(n)}$
$\implies \sum \frac{1}{\ln(n)}$ diverges.
A: You already have the radio that is 4, so
$$|x|<4$$
Result in
$$-4<x<4$$
Then you have the interval $(-4,4)$, and now you need to evaluate in the serie to know if the interval would change in order to be open or close.
Evaluate for $x=-4$ $$\sum_{n=0}^\infty {(-1)^n\cdot(-4)^n\over 4^n\cdot ln(n)} \qquad Diverges$$
And evaluate for $x=4$ $$\sum_{n=0}^\infty {(-1)^n\cdot(4)^n\over 4^n\cdot ln(n)} \qquad Converges$$
Now, the intervals for the serie are $(-4,4]$
