Answer
converges
Work Step by Step
Given $$\sum_{n=3}^{\infty} \frac{1}{n(\ln n)^{2}-n}$$ Since $$\sum_{n=3}^{\infty} \frac{1}{n(\ln n)^{2}}$$ By using the integral test, since $f(x)= \frac{1}{x(\ln x)^2}$ is decreasing and \begin{align*} \int_{3}^{\infty} \frac{1}{x(\ln x)^{2}} d x&= \frac{-1}{\ln x}\bigg|_{3}^{\infty}\\ &=\frac{1}{\ln 3}\lt \infty
\end{align*}
then $ \sum_{n=3}^{\infty} \frac{1}{n(\ln n)^{2}}$ converges and
\begin{align*}
\lim_{n\to\infty} \frac{a_n}{b_n}&= \lim_{n\to\infty} \frac{n(\ln n)^{2}}{n(\ln n)^{2}-n}\\
&= \lim_{n\to\infty} \frac{1}{1-\frac{1}{(\ln n)^2}}\\
&=1
\end{align*}
Hence, the given series also converges.
