Chapter 12.4, Problem 57E

To determine

To find the limit of the sequence $a_n=\frac{16}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]$ and verify the limit numerically by using a graphing utility to complete the table

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$a_n$

Answer to Problem 57E

  $\underset{n\to \infty }{\lim }{a}_n=\frac{16}{3}\simeq 5.3333$

Explanation of Solution

Given information :

Given sequence $a_n=\frac{16}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]$ and to verify the limit numerically given table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$a_n$

To find the limit of $a_n=\frac{16}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]$ by algebraically:

  $\begin{array}{l}.i.e,\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{16}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]\\ =\underset{n\to \infty }{\lim }\frac{16}{n^3}\left[\frac{n^3\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right]\end{array}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\left[\frac{16\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{6}\right]\\ =\frac{16\left(1+0\right)\left(2+0\right)}{6}\left(\because \underset{n\to \pm \infty }{\lim }\frac{1}{n}=0\right)\\ \therefore \underset{n\to \infty }{\lim }{a}_n=\frac{16}{3}\simeq 5.3333\end{array}$

To verify the result numerically as we have the following table,

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$a_n$	$16$	$6.16$	$5.4136$	$5.3413$	$5.3341$	$5.33341$	$5.333341$

From above table, we observe that the sequence $a_n$ approaches $5.333341$ as $n$ becomes a large number.

Hence, $\underset{n\to \infty }{\lim }{a}_n=\frac{16}{3}\simeq 5.3333$