Chapter 9.1, Problem 75E

To determine

To prove: To prove that $L_1\ne L_2$.

Explanation of Solution

Given:

  $\underset{n\to \infty }{\lim }{a}_n=L_1,\underset{m\to \infty }{\lim }{a}_m=L_2,$

Calculation:

Given that $\epsilon >0$ , By definition of convergence, for $\epsilon$ there corresponds N such that for all

  $\begin{array}{l}k(n)>N_1.\\ \left|a_n-L_1\right|<\epsilon \end{array}$

Assume $a_n$ converges.

Let $N=\max \{N_1,N_2\}$

Then, for $n>N$ , $\left|a_n-L_1\right|<\epsilon$ and $\left|a_{i(n)}-L_2\right|<\epsilon$ for infinitely many $n$ . This implies $\underset{n\to \infty }{\lim }{a}_n=L_1,\underset{m\to \infty }{\lim }{a}_m=L_2,$ .But $L_1\ne L_2$ Since the limit of a sequence is unique. Therefore, $a_n$ does not converge and hence diverge.