Chapter 11.4, Problem 56E

To determine

To calculate:

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

Answer to Problem 56E

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

Explanation of Solution

Given information:

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

  $a_n=\frac{3n\left(n+1\right)}{n^2}-\frac{4}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2$

Calculation:

For completing the table of the sequence,

Consider the given functions,

  $a_n=\frac{3n\left(n+1\right)}{n^2}-\frac{4}{n^4}{\left[\frac{n\left(n+1\right)}{2}\right]}^2$

Or

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

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

  $\begin{array}{l}{a}_{{10}^0}=\frac{3\left(1\right)\left(1+1\right)}{1}-\frac{4}{1^4}{\left[\frac{\left(1\right)\left(1+1\right)}{2}\right]}^2\\ =6-4\\ =2\end{array}$

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

  $\begin{array}{l}{a}_{{10}^1}=\frac{2{\left(10\right)}^{2}5\left(10\right)-1}{{10}^2}\\ =\frac{200+50-1}{100}\\ =\frac{249}{100}\\ =2.49\end{array}$

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

  $\begin{array}{l}{a}_{{10}^2}=\frac{2{\left(100\right)}^{2}5\left(100\right)-1}{{100}^2}\\ =\frac{20000+500-1}{10000}\\ =2.0499\end{array}$

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

  $\begin{array}{l}{a}_{{10}^3}=\frac{2{\left(1000\right)}^{2}5\left(1000\right)-1}{{1000}^2}\\ =\frac{2000000+5000-1}{1000000}\\ =2.00499\end{array}$

So the required table of the sequence is:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$
$a_n$	$2$	$2.49$	$2.049$	$2.0049$

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

Since, the sequence $a_n$ is,

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

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left[2+\frac{5}{n}-\frac{1}{n^2}\right]\\ =2+0+0\\ =2\end{array}$

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