Chapter 13.4, Problem 29E

To determine

To find : If the sequence is convergent, find its limit, if it divergent explain why.

Answer to Problem 29E

The limit does not exist so the sequence is divergent.

Explanation of Solution

Given information:

The given sequence is $a_n=\sin \left(n\pi /2\right)$

Calculation :

A sequence $a_1$ , $a_2$ , $a_3$ …… has the limit $L$ and we write

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

if the nth term $a_n$ of the sequence can be made arbitrarily close to $L$ by taking $n$ sufficiently large. If $\underset{n\to \infty }{\lim }{a}_n$ exists, then the sequences converges otherwise it diverges.

Find $\underset{n\to \infty }{\lim }{a}_n$.

  $\underset{n\to \infty }{\lim }\sin \left(n\pi /2\right)$

Write the first few terms of the sequence:

  $a_1=\sin \left(\frac{\pi }{2}\right)=1$

  $a_2=\sin \left(\frac{2\pi }{2}\right)=0$

  $a_3=\sin \left(\frac{3\pi }{2}\right)=-1$

  $a_4=\sin \left(\frac{4\pi }{2}\right)=0$ and so on.

It can be observed that as the value of $n$ increases, the value of $\sin \left(n\pi /2\right)$ oscillates between $-1$ , 0 and 1 due to periodic property of sine function. So as the value of $n$ is sufficiently large the value of $\sin \left(n\pi /2\right)$ does not approach any certain number. Therefore,

  $\underset{n\to \infty }{\lim }\sin \left(n\pi /2\right)$ does not exist.

Hence,

The limit does not exist so the sequence is divergent.