Chapter 13, Problem 41RE

To determine

To determine whether the sequence is convergent or divergent and then explain if divergent or find limit if convergent

Answer to Problem 41RE

No limit exists for this sequence.

Explanation of Solution

Given information:

Consider the given information,

  $a_n=\cos \left(\frac{n\pi }{2}\right)$

CALCULATION:

To find the sequence as,

  $\begin{array}{l}{a}_n=\cos \left(\frac{n\pi }{2}\right)\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\cos \left(\frac{n\pi }{2}\right)\end{array}$

Since, $n$ is a positive integer?

  $\begin{array}{l}4n>n\text{ and }4n+2>n\\ f\left(4n\right)=\cos \left(2n\pi \right)=1\\ \text{And }f\left(4n+2\right)=\cos \left(\left(2n+1\right)\pi \right)=-1\\ \text{Talking }\in =\frac{1}{2},\\ \left|1-L\right|<\frac{1}{2}\text{ and }\left|-1-L\right|<\frac{1}{2}\\ =\left|1+L\right|<\frac{1}{2}\end{array}$

Therefore, $\left|1-L\right|+\left|1+L\right|<1$

By the triangle inequality,

  $\begin{array}{l}\left|1-L+1+L\right|<1\\ =2<1\end{array}$

Therefore, it is a contradiction. Hence there is no such limit $L$