Chapter B, Problem 22AYU

To determine

To find: Determine whether each sequence converges or diverges. If it converges, find its limit.

Answer to Problem 22AYU

Solution:

The given sequence converges. The limit is 2.

Explanation of Solution

Given:

$\left\{b_n\right\} = \left\{\frac{2n^2 + {\left(-1\right)}^n}{n^2}\right\}$

Calculation:

$\left\{b_n\right\} = \left\{\frac{2n^2 + {\left(-1\right)}^n}{n^2}\right\} = \frac{2n^2}{n^2} + \frac{{\left(-1\right)}^n}{n^2}$

Hence the sequence converges. Therefore we will find the limit

$\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\text{lim}}\left\{\frac{2n^2+{\left(-1\right)}^n}{n^2}\right\}=\underset{n\to \infty }{\text{lim}}\left\{\frac{2n^2}{n^2}+\frac{{\left(-1\right)}^n}{n^2}\right\}=2$