Chapter 8, Problem 3RQ

To determine

To find: The statement is true or false and if the statement then given the explanation.

Answer to Problem 3RQ

The $\underset{n\to \infty }{\lim }{a}_{2n+1}=L$ is true as there is a positive integer $m$ such that $\left|a_n-L\right|<\epsilon$ for all the value of $n\ge m$ and as the value of $n\ge m$ .

Explanation of Solution

Given:

The given statement is that if $\underset{n\to \infty }{\lim }{a}_n=L$ .Then $\underset{n\to \infty }{\lim }{a}_{2n+1}=L$ .

Calculation:

Consider for the given $\epsilon >0$ there is a positive integer $m$ such that $\left|a_n-L\right|<\epsilon$ for all the value of $n\ge m$ and as the value of $n\ge m$ .

Then,

  $2n+1\ge m$

Thus, $\left|a_{2n+1}-L\right|<\epsilon$ for all $\left(2n+1\right)\ge m$

Thus, the $\underset{n\to \infty }{\lim }{a}_{2n+1}=L$ is true as there is a positive integer $m$ such that $\left|a_n-L\right|<\epsilon$ for all the value of $n\ge m$ and as the value of $n\ge m$ .