Chapter 11.4, Problem 53E

To determine

To calculate:

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

Answer to Problem 53E

Limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=\frac{3}{2}$ .

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$2$	$1.55$	$1.505$	$1.5005$

Explanation of Solution

Given information:

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

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

Calculation:

For completing the table of the sequence,

Consider the given functions,

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

Or

  $\begin{array}{l}{a}_n=\frac{1}{n}\left(n+\frac{1}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(\frac{1}{n}.n+\frac{1}{n}.\frac{1}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(1+\frac{1}{n^2}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\left(1+\frac{1}{n^2}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =1+\frac{n^2+n}{2n^2}\\ =1+\frac{n+1}{2n}\\ =\frac{3n+1}{2n}\end{array}$

Or

  $a_n=\frac{3n+1}{2n}$

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

  $\begin{array}{l}{a}_1=\frac{3\left(1\right)+1}{2\left(1\right)}\\ =\frac{4}{2}\\ =2\end{array}$

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

  $\begin{array}{l}{a}_{10}=\frac{3\left(10\right)+1}{2\left(10\right)}\\ =\frac{31}{20}\\ =1.55\end{array}$

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

  $\begin{array}{l}{a}_{{10}^2}=\frac{3\left({10}^2\right)+1}{2\left(10\right)}\\ =\frac{301}{200}\\ =1.505\end{array}$

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

  $\begin{array}{l}{a}_{{10}^3}=\frac{3\left({10}^3\right)+1}{2{\left(10\right)}^3}\\ =\frac{3001}{2000}\\ =1.5005\end{array}$

So the required table of the sequence is:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$2$	$1.55$	$1.505$	$1.5005$

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

Since, the sequence $a_n$ is,

  $\begin{array}{l}{a}_n=\frac{1}{n}\left(n+\frac{1}{n}\left[\frac{n\left(n+1\right)}{2}\right]\right)\\ =\frac{3n+1}{2n}\end{array}$

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{3n+1}{2n}\\ =\underset{n\to \infty }{\lim }\frac{3n\left(1+\frac{1}{3n}\right)}{2n}\\ =\underset{n\to \infty }{\lim }\frac{3\left(1+\frac{1}{3n}\right)}{2}\\ =\frac{3\left(1+\frac{8}{3.\infty }\right)}{2}\\ =\frac{3\left(1+0\right)}{2}\\ =\frac{3}{2}\end{array}$

Hence, the required limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=\frac{3}{2}$ .