Chapter 11.4, Problem 52E

To determine

To calculate:

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

Answer to Problem 52E

First five terms are $1,-\frac{1}{4},\frac{1}{9},-\frac{1}{16},\frac{1}{25}$ and limit of the sequence is $0$ .

Explanation of Solution

Given information:

  $a_n=\frac{{\left(-1\right)}^{n+1}}{n^2}$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{{\left(-1\right)}^{n+1}}{n^2}$

First five terms are : $a_1,a_2,a_3,a_4,a_5$

For $n=1$ :

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

For $n=2$ :

  $\begin{array}{l}{a}_n=\frac{{\left(-1\right)}^{2+1}}{2^2}\\ =-\frac{1}{4}\end{array}$

For $n=3$ :

  $\begin{array}{l}{a}_n=\frac{{\left(-1\right)}^{3+1}}{3^2}\\ =\frac{1}{9}\end{array}$

For $n=4$ :

  $\begin{array}{l}{a}_n=\frac{{\left(-1\right)}^{4+1}}{4^2}\\ =-\frac{1}{16}\end{array}$

For $n=5$ :

  $\begin{array}{l}{a}_n=\frac{{\left(-1\right)}^{5+1}}{5^2}\\ =\frac{1}{25}\end{array}$

Now for finding out the limit of the sequence in case it exists:

  $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^{n+1}}{n^2}$

Use the property of limit,

  $\underset{n\to \infty }{\lim }f\left(x\right).g\left(x\right)=\underset{n\to \infty }{\lim }f\left(x\right).\underset{n\to \infty }{\lim }g\left(x\right)$

Thus,

  $\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^{n+1}}{n^2}=\underset{n\to \infty }{\lim }{\left(-1\right)}^{n+1}.\underset{n\to \infty }{\lim }\frac{1}{n^2}$

The limit $\underset{n\to \infty }{\lim }\frac{1}{n^2}$ is $0$ as the degree of the numerator is less than the degree of the denominator.

Thus, the limit of the sequence is,

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^{n+1}.\underset{n\to \infty }{\lim }\frac{1}{n^2}\\ =\underset{n\to \infty }{\lim }{\left(-1\right)}^{n+1}.0\\ =0\end{array}$