Chapter 12.4, Problem 58E

To determine

To find the limit of the sequence $a_n=\frac{n(n+1)}{n^2}-\frac{1}{n^4}{\left[\frac{n(n+1)}{2}\right]}^2$ 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 58E

  $\underset{n\to \infty }{\lim }{a}_n=\frac{3}{4}\simeq 0.75$

Explanation of Solution

Given information :

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

  $\begin{array}{l}.i.e,\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(\frac{n(n+1)}{n^2}-\frac{1}{n^4}{\left[\frac{n(n+1)}{2}\right]}^2\right)\\ =\underset{n\to \infty }{\lim }\frac{n(n+1)}{n^2}-\underset{n\to \infty }{\lim }\frac{1}{n^4}{\left[\frac{n(n+1)}{2}\right]}^2\end{array}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{n^2\left(1+\frac{1}{n}\right)}{n^2}-\underset{n\to \infty }{\lim }\frac{n^4}{n^4}{\left[\frac{\left(1+\frac{1}{n}\right)}{2}\right]}^2\\ =\underset{n\to \infty }{\lim }\frac{\left(1+\frac{1}{n}\right)}{1}-\underset{n\to \infty }{\lim }{\left[\frac{\left(1+\frac{1}{n}\right)}{2}\right]}^2\\ =\left(1+0\right)-{\left[\frac{\left(1+0\right)}{2}\right]}^2\left(\because \underset{n\to \pm \infty }{\lim }\frac{1}{n}=0\right)\\ =1-\frac{1}{4}\\ \therefore \underset{n\to \infty }{\lim }{a}_n=\frac{3}{4}\simeq 0.75\end{array}$

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

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$	${10}^5$	${10}^6$
$a_n$	$1$	$0.798$	$0.754975$	$0.750499$	$0.75005$	$0.750005$	$0.7500005$

From above table, we observe that the sequence $a_n$ approaches $0.75$ as $n$ becomes a large number.

Hence, $\underset{n\to \infty }{\lim }{a}_n=\frac{3}{4}\simeq 0.75$