Chapter 11.4, Problem 54E

To determine

To calculate:

The limit of the sequence as $n$ approaches infinity and complete the table.

Answer to Problem 54E

Limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=12$ .

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$20$	$12.8$	$12.08$	$12.008$

Explanation of Solution

Given information:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$

  $a_n=\frac{4}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{4}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)$

Or

  $\begin{array}{l}{a}_n=\frac{4}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(\frac{4}{n}.n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(\frac{4}{n}.n+\frac{4}{n}.\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(4+\frac{16}{n^2}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(4+\frac{\left(8n+8\right)}{n}\right)\\ =\frac{12n+8}{n}\end{array}$

Or

  $a_n=\frac{12n+8}{n}$

Then the value of the sequence $a_n$ at $n={10}^0=1$ .

  $\begin{array}{l}{a}_1=\frac{12\left(1\right)+8}{1}\\ =20\end{array}$

Then the value of the sequence $a_n$ at $n=10$ .

  $\begin{array}{l}{a}_{{10}^1}=\frac{12\left(10\right)+8}{10}\\ =\frac{128}{10}\\ =12.8\end{array}$

Then the value of the sequence $a_n$ at $n={10}^2$ .

  $\begin{array}{l}{a}_{{10}^1}=\frac{12\left(100\right)+8}{100}\\ =\frac{1208}{100}\\ =12.08\end{array}$

Then the value of the sequence $a_n$ at $n={10}^3$ .

  $\begin{array}{l}{a}_{{10}^1}=\frac{12\left(1000\right)+8}{1000}\\ =\frac{12008}{1000}\\ =12.008\end{array}$

So the required table of the sequence is:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$20$	$12.8$	$12.08$	$12.008$

Now to finding out the limit of the sequence $a_n$ as $n$ approaches to infinity.

Since, the sequence $a_n$ is,

  $\begin{array}{l}{a}_n=\frac{4}{n}\left(n+\frac{4}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\frac{12n+8}{n}\end{array}$

Then, limit of the sequence $a_n$ as $n$ approaches to infinity.

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left[\frac{12n+8}{n}\right]\\ =\underset{n\to \infty }{\lim }\frac{12n\left(1+\frac{8}{12n}\right)}{n}\\ =\underset{n\to \infty }{\lim }\frac{12n\left(1+\frac{8}{12n}\right)}{1}\\ =\frac{12\left(1+\frac{8}{12.\infty }\right)}{1}\\ =12\end{array}$

Hence, the required limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=12$ .