Answer
$$ -5 \leq x\leq-3 $$
Work Step by Step
Given $$\sum_{n=2}^{\infty} \frac{(x+4)^{n}}{(n \ln n)^{2}} $$ Since $a_n = \frac{(x+4)^{n}}{(n \ln n)^{2}} $ and $a_{n+1} = \frac{(x+4)^{n+1}}{((n+1) \ln (n+1))^{2}} $, then \begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{(x+4)^{n+1}}{((n+1) \ln (n+1))^{2}} \frac{(n \ln n)^{2}}{(x+4)^{n}} \right|\\ &=|(x+4)| \lim _{n \rightarrow \infty}\left(\frac{n}{n+1}\right)^{2} \lim _{n \rightarrow \infty}\left(\frac{\ln n}{\ln (n+1)}\right)^{2}\\ &= |(x+4)| \left( \lim _{n \rightarrow \infty}\frac{n}{n+1}\right)^{2}\left( \lim _{n \rightarrow \infty}\frac{\ln n}{\ln (n+1)}\right)^{2}\\ &= |(x+4)| \left( \lim _{n \rightarrow \infty}\frac{1/n}{1/(n+1)}\right)^{2}\\ &= |(x+4)| \end{aligned} Then the series converges for $$ |x+4|\lt 1 \ \to -5 \lt x\lt-3 $$
Now, we check the end points.
For $x= -5 $ $$\sum_{n=2}^{\infty} \frac{(x+4)^{n}}{(n \ln n)^{2}}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(n \ln n)^{2}} $$
which is an alternating series with $\lim_{n\to \infty} a_n =0 $, which converges.
For $x= -3$ $$\sum_{n=2}^{\infty} \frac{(x+4)^{n}}{(n \ln n)^{2}}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(n \ln n)^{2}} $$ which is convergent by the integral test.
Hence, the interval of convergence is $$ -5 \leq x\leq-3 $$
