Chapter 9.1, Problem 43E

To determine

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

Answer to Problem 43E

The sequence converges to $0$ .

Explanation of Solution

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

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 $\underset{x\to x_0}{\lim }f\left(x\right)=L$

$n!$ is greater than or equal to $n$ for all $n\ge 1$ .

Therefore $n!\ge n$ for all $n\ge 1$

  $\Rightarrow \frac{1}{n!}\le \frac{1}{n}$

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

  $0<\frac{1}{2^n}\le \frac{1}{n}$

  $\Rightarrow \underset{n\to \infty }{\lim }0<\underset{n\to \infty }{\lim }\frac{1}{n!}\le \underset{n\to \infty }{\lim }\frac{1}{n}$

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

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

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