Chapter 9.1, Problem 40E

To determine

To Calculate: The limit of a sequence if it converges.

Answer to Problem 40E

The sequence does not converge.

Explanation of Solution

Given Information: The sequence is defined as $a_n=\cos \left(n\frac{\pi }{2}\right)$

Concept Used:

• A sequence converges if and only if the limit is finite and single valued.
• $\cos \left(2k\pi +\theta \right)=\cos \left(\theta \right)$ , for all integer values of $p$ .

Let $n=4p$ , where $p$ is a positive integer.

Then

  $a_n=\cos \left(N\cdot \frac{\pi }{2}\right)=\cos \left(4p\cdot \frac{\pi }{2}\right)=\cos \left(2p\pi \right)=\cos \left(0\right)=1$

  $a_{n+1}=\cos \left[\left(N+1\right)\cdot \frac{\pi }{2}\right]=\cos \left[\left(4p+1\right)\cdot \frac{\pi }{2}\right]=\cos \left(2p\pi +\frac{\pi }{2}\right)=\cos \left(\frac{\pi }{2}\right)=0$

  $a_{n+2}=\cos \left[\left(N+2\right)\cdot \frac{\pi }{2}\right]=\cos \left[\left(4p+2\right)\cdot \frac{\pi }{2}\right]=\cos \left(2p\pi +\pi \right)=\cos \left(\pi \right)=-1$

  $a_{n+3}=\cos \left[\left(N+3\right)\cdot \frac{\pi }{2}\right]=\cos \left[\left(4p+3\right)\cdot \frac{\pi }{2}\right]=\cos \left(2p\pi +\frac{3\pi }{2}\right)=\cos \left(\frac{3\pi }{2}\right)=0$

  $a_{n+4}=a_n$

  $a_{n+5}=a_1$

  $⋮$

So, however large $N$ may be, no $4$ consecutive terms have the same value. Therefore, the sequence does not converge.