Chapter 11, Problem 89RE

To determine

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

Answer to Problem 89RE

  $-1$, $\frac{1}{8}$, $-\frac{1}{27}$, $\frac{1}{64}$, $-\frac{1}{125}$

And

  $\underset{n\to \infty }{\lim }{a}_n=0$

Explanation of Solution

Given information:

The sequence is

  $a_n-\frac{{\left(-1\right)}^n}{n^3}$

Calculation:

Consider the following sequence

  $a_n-\frac{{\left(-1\right)}^n}{n^3},n\ge 1$

The first five terms of the sequence are $a_1,a_2,a_3,a_4,a_5.$

  $\begin{array}{c}{a}_1=\frac{{\left(-1\right)}^1}{1^3}\\ =-1\\ a_2=\frac{{\left(-1\right)}^2}{2^3}\\ =\frac{1}{8}\\ a_3=\frac{{\left(-1\right)}^3}{3^3}\\ =-\frac{1}{27}\\ a_4=\frac{{\left(-1\right)}^4}{4^3}\\ =\frac{1}{64}\\ a_5=\frac{{\left(-1\right)}^5}{5^3}\\ =-\frac{1}{125}\end{array}$

Find the limit of the sequence, if it exists.

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^n}{n^3}\\ =\underset{n\to \infty }{\lim }{\left(-1\right)}^n\cdot \underset{n\to \infty }{\lim }\frac{1}{n^3}\end{array}$

The limit $\underset{n\to \infty }{\lim }\frac{1}{n^3}$ is 0 as the degree of the numerator is less than the degree of the denominator.

Therefore, the limit of the sequence is

  $\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\cdot \underset{n\to \infty }{\lim }\frac{1}{n^3}\\ =\underset{n\to \infty }{\lim }{\left(-1\right)}^n\cdot 0\\ =0\end{array}$