Chapter 12, Problem 91RE

To determine

To write: the first five terms of the sequence and the limit of the sequence.

Answer to Problem 91RE

The first five terms of the sequence are

  $\frac{1}{5},\frac{1}{2},\frac{9}{11},\frac{8}{7},\frac{25}{17}$

And

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

Explanation of Solution

Given information:

The sequence is

  $a_n-\frac{n^2}{3n+2}$

Calculation:

Consider the following sequence

  $a_n-\frac{n^2}{3n+2}$

Substitute $n=1$ in the expression $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{c}{a}_1=\frac{1^2}{3\times 1+2}\\ =\frac{1}{5}\end{array}$

Substitute $n=2$ in the expression $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{c}{a}_1=\frac{2^2}{3\times 2+2}\\ =\frac{1}{2}\end{array}$

Substitute $n=3$ in the expression $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{c}{a}_1=\frac{3^2}{3\times 3+2}\\ =\frac{9}{11}\end{array}$

Substitute $n=4$ in the expression $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{c}{a}_1=\frac{4^2}{3\times 4+2}\\ =\frac{8}{7}\end{array}$

Substitute $n=5$ in the expression $a_n=\frac{n^2}{3n+2}$

  $\begin{array}{c}{a}_1=\frac{5^2}{3\times 5+2}\\ =\frac{25}{17}\end{array}$

Therefore the first five terms of the sequence are

  $\frac{1}{5},\frac{1}{2},\frac{9}{11},\frac{8}{7},\frac{25}{17}$

Limits of the sequence can be found as

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{n^2}{3n+2}\\ =\infty \end{array}$

The value of the limit goes unbounded. Therefore limit doesn’t exist.