Chapter B, Problem 12AYU

In Problems 5-28, determine whether each sequence converges or diverges. If it converges, find its limit.

$\left\{a_n\right\}=\left\{\frac{12n^2}{n^3-3n+1}\right\}$

To determine

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

Answer to Problem 12AYU

Solution:

The given sequence converges. The limit is 0.

Explanation of Solution

Given:

$\left\{a_n\right\} = \left\{\frac{12n^2}{n^3 - 3n + 1}\right\}$

Calculation:

$\left\{a_n\right\} = \left\{\frac{12n^2}{n^3 - 3n + 1}\right\}$

$\left\{\frac{12n^2}{n^3 - 3n + 1}\right\} = \left\{\frac{12n^2}{n^3\left(1 - \frac{3n}{n^3} + \frac{1}{n^3}\right)}\right\} = \left\{\frac{12}{n\left(1 - \frac{3n}{n^3} + \frac{1}{n^3}\right)}\right\}$

Hence the sequence converges. Therefore we will find the limit

$\underset{n \to \infty }{\lim } a_n = \underset{n \to \infty }{\lim }\left\{\frac{12}{n\left(1 - \frac{3n}{n^3} + \frac{1}{n^3}\right)}\right\} = 0$