Chapter 11.4, Problem 47E

To determine

To calculate:

The five terms and limit of the sequence, if limit does not exist then give reason.

Answer to Problem 47E

First five terms are $\frac{1}{5},\frac{1}{2},\frac{9}{11},\frac{8}{7},\frac{25}{17}$ and limit of the sequence does not exist.

Explanation of Solution

Given information:

  $a_n=\frac{n^2}{3n+2}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{n^2}{3n+2}$

Now substitute $1$ for $n$ in $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{l}{a}_n=\frac{{\left(1\right)}^2}{3\left(1\right)+2}\\ =\frac{1}{3+2}\\ =\frac{1}{5}\end{array}$

The first term of the sequence is $\frac{1}{5}$ .

Similarly substitute $2,3,4,5$ for $n$ in $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{l}{a}_2=\frac{{\left(2\right)}^2}{3\left(2\right)+2}=\frac{4}{6+2}=\frac{4}{8}=\frac{1}{2}\\ a_3=\frac{{\left(3\right)}^2}{3\left(3\right)+2}=\frac{9}{9+2}=\frac{9}{11}\\ a_4=\frac{{\left(4\right)}^2}{3\left(4\right)+2}=\frac{16}{12+2}=\frac{8}{7}\\ a_5=\frac{{\left(5\right)}^2}{3\left(5\right)+2}=\frac{25}{15+2}=\frac{25}{17}\end{array}$

First five terms are : $\frac{1}{5},\frac{1}{2},\frac{9}{11},\frac{8}{7},\frac{25}{17}$

In case $\left\{a_n\right\}$ is a sequence such that $f\left(n\right)=a_n$ for every positive integer $n$ then $\underset{n\to \infty }{\lim }{a}_n=L$ .

From the first five terms of the sequence It is observed that value of the $a_n$ increases without bound.

Thus, the limit does not exist.