Chapter 11.4, Problem 55E

To determine

To calculate:

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

Answer to Problem 55E

Limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=5$ .

required table of the sequence is:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$13.33$	$5.683$	$5.06683$	$5.0066683$

Explanation of Solution

Given information:

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

  $a_n=\frac{10}{n^3}\left[\frac{n\left(n+1\right)\left(3n+1\right)}{6}\right]$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{10}{n^3}\left[\frac{n\left(n+1\right)\left(3n+1\right)}{6}\right]$

Or

  $\begin{array}{l}{a}_n=\frac{10}{n^3}\left[\frac{n\left(n+1\right)\left(3n+1\right)}{6}\right]\\ =\frac{10}{n^3}\left[\frac{3n^3+4n^2+n}{6}\right]\\ =\left[5+\frac{20}{3n}+\frac{5}{3n^2}\right]\\ =\left[5+\frac{20n+5}{3n^2}\right]\end{array}$

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

  $\begin{array}{l}{a}_{{10}^0}=\left[5+\frac{20\left(1\right)+5}{3{\left(1\right)}^2}\right]\\ =\left[5+\frac{25}{3}\right]\\ =\frac{40}{3}\\ =13.33\end{array}$

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

  $\begin{array}{l}{a}_{{10}^1}=\left[5+\frac{20\left(10\right)+5}{3{\left(10\right)}^2}\right]\\ =\left[5+\frac{205}{300}\right]\\ =\frac{1705}{300}\\ =5.6833\end{array}$

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

  $\begin{array}{l}{a}_{{10}^2}=\left[5+\frac{20\left(100\right)+5}{3{\left(100\right)}^2}\right]\\ =\left[5+\frac{2005}{30000}\right]\\ =152005\end{array}$

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

  $\begin{array}{l}{a}_{{10}^2}=\left[5+\frac{20\left(1000\right)+5}{3{\left(1000\right)}^2}\right]\\ =\left[5+\frac{20005}{3000000}\right]\\ =\frac{15020005}{3000000}\\ =5.0066683\end{array}$

So the required table of the sequence is:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$13.33$	$5.683$	$5.06683$	$5.0066683$

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

Since, the sequence $a_n$ is,

  $\begin{array}{l}{a}_n=\frac{10}{n^3}\left[\frac{n\left(n+1\right)\left(3n+1\right)}{6}\right]\\ =\left[5+\frac{20n+5}{3n^2}\right]\\ =\left[5+\frac{20}{3n}+\frac{5}{3n^2}\right]\end{array}$

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left[5+\frac{20}{3n}-\frac{5}{3n^2}\right]\\ =5+0+0\\ =5\end{array}$

Hence, the required limit of the sequence $a_n$ as $n$ approaches to infinity is $\underset{n\to \infty }{\lim }{a}_n=5$ .