Chapter 10.1, Problem 75E

To determine

That the two sub sequences of the sequence $\left\{a_n\right\}$ have different limits.

Answer to Problem 75E

The $a_n$ does not converge and hence diverge.

Explanation of Solution

Given information:

The sequence {a_n} converges to the number L if to every positive number $\epsilon$ there corresponds an integer N such that for all n,

  $n>N\Rightarrow \left|a_n-L\right|<\epsilon$

Formula used:

  $k\left(n\right)>N_1,\left|a_{k\left(n\right)}-L_1\right|<\in$

Calculation:

Consider the two sub sequences $a_{k\left(n\right)}\text{and}{a}_{i\left(n\right)}$ .

Where, $\underset{n\to \infty }{\lim }{a}_{k\left(n\right)}=L_1\text{and}\underset{n\to \infty }{\lim }{a}_{i\left(n\right)}=L_2\left(L_1\ne L_2\right)$

Given $\in >0$

There corresponds an $N_1$ such that for $k\left(n\right)>N_1,\left|a_{k\left(n\right)}-L_1\right|<\in$ their corresponds an $N_2$ such that for $i\left(n\right)>N_2,\left|a_{i\left(n\right)}-L_2\right|<\in$

Assume $a_n$ converges. Let $N=\max \left\{N_1,N_2\right\}$ . Then for $n>N$ , that $\left|a_n-L_1\right|<\in$ and $\left|a_n-L_2\right|<\in$ for infinitely many n. This implies that $\underset{n\to \infty }{\lim }{a}_n=L_1$ and $\underset{n\to \infty }{\lim }{a}_n=L_2$ . But $L_1\ne L_2$ since the limit of a sequence is unique. Therefore, $a_n$ does not converge and hence diverge.

Conclusion:

  $a_n$ does not converge and hence diverge.