Chapter 8.1, Problem 32E

To determine

To find: To determine whether the sequence converges or diverges and find limit of converges.

Answer to Problem 32E

The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=0}$

Explanation of Solution

Given information: $a_n=\frac{{\left(\ln n\right)}^2}{n}$

Solution:

To determine converges or diverges we need to check the limit of the sequence of the partial sums.

  $\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{{\left(\ln n\right)}^2}{n}$

Applying L’ Hospital rule

  $\begin{array}{l}\Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{2\times \left(\ln n\right)\times \frac{1}{n}}{1}\left[\begin{array}{l}\because \frac{d}{dx}{\left(\ln n\right)}^2=2\ln n\\ \text{ and }\frac{d}{dx}\left(\ln n\right)=\frac{1}{n}\end{array}\right]\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{2\ln n}{n}\end{array}$

Again applying L’ Hospital rule

  $\begin{array}{l}\Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{\frac{2}{n}}{1}\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{2}{n}\\ \Rightarrow \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }0\end{array}$

  $\therefore$ Limit exists and is finite.

Hence, The series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ is converges and its value is ${\displaystyle \sum _{n=0}^{\infty }{a}_n=0}$