## Calculus (3rd Edition)

$$-5 \leq x\leq-3$$
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 $$-5 \leq x\leq-3$$