# Chapter 11 - Infinite Series - 11.3 Convergence of Series with Positive Terms - Exercises - Page 557: 64

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.

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