Chapter 8.1, Problem 4E

To determine

To Find: The first nine terms of the sequence $\left\{\cos \left(\frac{n\pi }{3}\right)\right\}$ and determine if the sequence appears to have a limit.

Answer to Problem 4E

The first nine term of the sequence is $\frac{1}{2},-\frac{1}{2},-1,-\frac{1}{2},\frac{1}{2},1,\frac{1}{2},-\frac{1}{2},1$ , as the sequence repeats itself it doesn’t appear to have any limits.

Explanation of Solution

Calculation:

Consider the given sequence is,

  $\left\{\cos \left(\frac{n\pi }{3}\right)\right\}$

The first term is,

  $\begin{array}{c}{a}_1=\left\{\cos \left(\frac{\left(1\right)\pi }{3}\right)\right\}\\ =\frac{1}{2}\end{array}$

The second term is,

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

The third term is,

  $\begin{array}{c}{a}_3=\left\{\cos \left(\frac{3\pi }{3}\right)\right\}\\ =-1\end{array}$

The fourth term is,

  $\begin{array}{c}{a}_4=\left\{\cos \left(\frac{4\pi }{3}\right)\right\}\\ =-\frac{1}{2}\end{array}$

The fifth term is,

  $\begin{array}{c}{a}_5=\left\{\cos \left(\frac{5\pi }{3}\right)\right\}\\ =\frac{1}{2}\end{array}$

The sixth term is,

  $\begin{array}{c}{a}_6=\left\{\cos \left(\frac{6\pi }{3}\right)\right\}\\ =1\end{array}$

The seventh term is,

  $\begin{array}{c}{a}_7=\left\{\cos \left(\frac{7\pi }{3}\right)\right\}\\ =\frac{1}{2}\end{array}$

The eight term is,

  $\begin{array}{c}{a}_8=\left\{\cos \left(\frac{8\pi }{3}\right)\right\}\\ =-\frac{1}{2}\end{array}$

The ninth term is,

  $\begin{array}{c}{a}_9=\left\{\cos \left(\frac{9\pi }{3}\right)\right\}\\ =-1\end{array}$

The first nine term of the sequence is $\frac{1}{2},-\frac{1}{2},-1,-\frac{1}{2},\frac{1}{2},1,\frac{1}{2},-\frac{1}{2},1$ , as the sequence repeats itself it doesn’t appear to have any limits.