Chapter 13.4, Problem 25E

To determine

To find : If the sequence is convergent, find its limit, if it divergent explain why.

Answer to Problem 25E

The limit does not exist so the sequence is divergent.

Explanation of Solution

Given information:

The given sequence is $a_n=\frac{n^2}{n+1}$

Calculation :

A sequence $a_1$ , $a_2$ , $a_3$ …… has the limit $L$ and we write

  $\underset{n\to \infty }{\lim }{a}_n=L$

if the nth term $a_n$ of the sequence can be made arbitrarily close to $L$ by taking $n$ sufficiently large. If $\underset{n\to \infty }{\lim }{a}_n$ exists, then the sequences converges otherwise it diverges.

Find $\underset{n\to \infty }{\lim }{a}_n$.

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{n^2}{n+1}=\underset{n\to \infty }{\lim }\frac{\frac{n^2}{n^2}}{\frac{n}{n^2}+\frac{1}{n^2}}\text{ (Divide numerator and denominator by }{n}^2\text{)}\\ \underset{n\to \infty }{\lim }\frac{n^2}{n+1}=\underset{n\to \infty }{\lim }\frac{1}{\frac{1}{n}+\frac{1}{n^2}}\\ \underset{n\to \infty }{\lim }\frac{n^2}{n+1}=\frac{1}{\frac{1}{\infty }+\frac{1}{{\infty }^2}}\\ \underset{n\to \infty }{\lim }\frac{n^2}{n+1}=\frac{1}{0+0}\text{ (Since }\frac{a}{\infty }=0\text{)}\\ \underset{n\to \infty }{\lim }\frac{n^2}{n+1}=\infty \end{array}$

The limit does not exist so the sequence is divergent.

Hence,

The limit does not exist so the sequence is divergent.