Chapter 9.1, Problem 42E

To determine

To Show: That the given sequence converges to a limit, using Sandwich Theorem.

Answer to Problem 42E

The sequence converges to $0$ .

Explanation of Solution

Given Information: The $n^{th}$ term of the sequence is given by $a_n=\frac{1}{2^n}$

Concept Used:Sandwich Theorem

If $f_1\left(x\right)\le f\left(x\right)\le f_2\left(x\right)$ in an interval $I$ and if $\underset{x\to x_0}{\lim }{f}_1\left(x\right)=\underset{x\to x_0}{\lim }{f}_2\left(x\right)=L$ , where $x_0\in I$ , then $\underset{x\to x_0}{\lim }f\left(x\right)=L$

$2^n$ is greater than $n$ for all $n\ge 1$ . For example,

  $2^1=2>1$

  $2^2=4>2$

  $2^3=8>3$

  $2^4=16>4$

  $⋮$

Therefore $2^n>n$ for all $n\ge 1$

  $\Rightarrow \frac{1}{2^n}<\frac{1}{n}$

And $\frac{1}{2^n}$ is always positive for all values of $n$ . So

  $0<\frac{1}{2^n}<\frac{1}{n}$

  $\Rightarrow \underset{n\to \infty }{\lim }0<\underset{n\to \infty }{\lim }\frac{1}{2^n}<\underset{n\to \infty }{\lim }\frac{1}{n}$

  $\Rightarrow 0<\underset{n\to \infty }{\lim }\frac{1}{2^n}<0$ (Since $\underset{n\to \infty }{\lim }\frac{1}{n}=0$ )

  $\Rightarrow \underset{n\to \infty }{\lim }\frac{1}{2^n}=0$

Therefore, the given sequence converges to the limit $0$ .