Answer
The radius of convergence: $4$
The interval of convergence: $(-4,4]$
Work Step by Step
Let $a_n=(-1)^n\frac{x^n}{4^n\ln n}$.
It can be checked that $\frac{a_{n+1}}{a_n}=\frac{-x\ln (n+1)}{4\ln n}$.
We know that $\lim_{n\to \infty}\frac{\ln (n+1)}{\ln n}=1$.
Then,
$\lim |\frac{a_{n+1}}{a_n}|=\lim \frac{-x\ln (n+1)}{4\ln n}=|\frac{x}{-4}|$
Find the interval of convergence:
$|\frac{x}{-4}|<1$
$|x|<4$
For $x=-4$, the series would be $\sum_{n=2}^\infty \frac{1}{\ln n}$ and it is divergent.
For $x=4$, the series would be $\sum_{n=2}^\infty \frac{(-1)^n}{\ln n}$ and it is convergent.
Thus, the radius of convergence is $4$ and the interval of convergence is $(-4,4]$.
