Chapter 12, Problem 90RE

To determine

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

Answer to Problem 90RE

The first five terms of the sequence are

  $1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5}$

And

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

Explanation of Solution

Given information:

The sequence is

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

Calculation:

Consider the following sequence

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

Substitute $n=1$ in the expression $a_n=\frac{{\left(-1\right)}^{n+1}}{n}$

  $\begin{array}{c}{a}_1=\frac{{\left(-1\right)}^{1+1}}{1}\\ =\frac{{\left(-1\right)}^2}{1}\\ =1\end{array}$

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

  $\begin{array}{c}{a}_2=\frac{{\left(-1\right)}^{2+1}}{2}\\ =\frac{{\left(-1\right)}^3}{2}\\ =-\frac{1}{2}\end{array}$

Substitute $n=3$ in the expression $a_n=\frac{{\left(-1\right)}^{n+1}}{n}$

  $\begin{array}{c}{a}_3=\frac{{\left(-1\right)}^{3+1}}{3}\\ =\frac{{\left(-1\right)}^4}{3}\\ =\frac{1}{3}\end{array}$

Substitute $n=4$ in the expression $a_n=\frac{{\left(-1\right)}^{n+1}}{n}$

  $\begin{array}{c}{a}_4=\frac{{\left(-1\right)}^{4+1}}{4}\\ =\frac{{\left(-1\right)}^5}{4}\\ =-\frac{1}{4}\end{array}$

Substitute $n=5$ in the expression $a_n=\frac{{\left(-1\right)}^{n+1}}{n}$

  $\begin{array}{c}{a}_5=\frac{{\left(-1\right)}^{5+1}}{5}\\ =\frac{{\left(-1\right)}^6}{5}\\ =\frac{1}{5}\end{array}$

Therefore the first five terms of the sequence are

  $1,-\frac{1}{2},\frac{1}{3},-\frac{1}{4},\frac{1}{5}$

Limits of the sequence can be found as

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

The value of the limit is found to be 0.