Chapter 8.1, Problem 14E

To determine

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

Answer to Problem 14E

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

Explanation of Solution

Given information: $a_n=\frac{n^3}{n+1}$

Solution:

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

  $\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{n^3}{n+1}$

Divide numerator and denominator by $\text{'}n\text{'}.$ we will get,

  $=\underset{n\to \infty }{\lim }\frac{n^2}{1+\frac{1}{n}}=\infty \left[\begin{array}{l}\because \underset{n\to \infty }{\lim }\frac{1}{n}=0\\ \Rightarrow \underset{n\to \infty }{\lim }{n}^2=\infty \end{array}\right]$

  $\therefore$ Limit exists and is infinite.

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