Chapter 9.PS, Problem 1PS

Decreasing Sequence Consider the sequence $a_n=\frac{n+1}{n^2+1}$,

(a) Use a graphing utility to graph the first $10$ terms of the sequence.

(b) Use the graph from part (a) to estimate the value of $a_n$ as $n$ approaches infinity.

(c) Complete the table.

$n$	$1$	$10$	$100$	$1000$	$10000$
$a_n$

(d) Use the table from part (c) to determine (if possible) the value of $a_n$ as $n$ approaches infinity.

To determine

(a)

To find:

The first $10$ term of the sequence by using graphing utility.

Answer to Problem 1PS

Solution:

The first $10$ terms of the given sequence are $1,0.6,0.4,0.294,0.230,$ $0.189,0.160,0.138,0.121$ $\text{and}0.108$.

Explanation of Solution

Given:

$a_n=\frac{\left(n+1\right)}{\left(n^2+1\right)}\mathrm{....}\left(1\right)$

Approach:

If the given series is in the form of $a_n$, then the first terms of the series are $a_1,a_2,a_3,a_4,\mathrm{...}$ and so on.

Calculation:

Substitute $1$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_1& =& \frac{\left(1+1\right)}{\left(1^2+1\right)}\\ & =& \frac{2}{2}\\ & =& 1\end{array}$

Substitute $2$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_2& =& \frac{\left(2+1\right)}{\left(2^2+1\right)}\\ & =& \frac{3}{5}\\ & =& 0.6\end{array}$

Substitute $3$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_3& =& \frac{\left(3+1\right)}{\left(3^2+1\right)}\\ & =& \frac{4}{10}\\ & =& 0.4\end{array}$

Substitute $4$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_4& =& \frac{\left(4+1\right)}{\left(4^2+1\right)}\\ & =& \frac{5}{17}\\ & =& 0.294\end{array}$

Substitute $5$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_5& =& \frac{\left(5+1\right)}{\left(5^2+1\right)}\\ & =& \frac{6}{26}\\ & =& 0.230\end{array}$

Substitute $6$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_6& =& \frac{\left(6+1\right)}{\left(6^2+1\right)}\\ & =& \frac{7}{37}\\ & =& 0.189\end{array}$

Substitute $7$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_7& =& \frac{\left(7+1\right)}{\left(7^2+1\right)}\\ & =& \frac{8}{50}\\ & =& 0.160\end{array}$

Substitute $8$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_8& =& \frac{\left(8+1\right)}{\left(8^2+1\right)}\\ & =& \frac{9}{65}\\ & =& 0.138\end{array}$

Substitute $9$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_9& =& \frac{\left(9+1\right)}{\left(9^2+1\right)}\\ & =& \frac{10}{82}\\ & =& 0.121\end{array}$

Substitute $10$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{10}& =& \frac{\left(10+1\right)}{\left({10}^2+1\right)}\\ & =& \frac{11}{101}\\ & =& 0.108\end{array}$

From the above calculation first $10$ terms of the sequence are $1,0.6,0.4,0.294,0.230,$ $0.189,0.160,0.138,0.121$ $\text{and}0.108$.

Table-1

$n$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$a_n$	$1$	$0.6$	$0.4$	$0.294$	$0.230$	$0.189$	$0.160$	$0.138$	$0.121$	$0.108$

The following figure represents the graph of the sequence $a_n=\frac{\left(n+1\right)}{\left(n^2+1\right)}$ by using table-1.

Figure (1)

To determine

(b)

To find:

The value of $a_n$, if $n$ is $\infty$.

Answer to Problem 1PS

Solution:

$0$

Explanation of Solution

Calculation:

Substitute $\infty$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{\infty }& =& \frac{\left(\infty +1\right)}{\left({\infty }^2+1\right)}\\ & =& \frac{\infty }{\infty }\\ & =& 0\end{array}$

From above calculation the value of $a_n$ is $0$.

Conclusion:

Hence, the value of $a_n$ is $0$ if $n$ is $\infty$.

To determine

(c)

To find:

The terms $a_1$, $a_{10}$, $a_{100}$, $a_{1000}$ and $a_{10000}$ and complete the table.

Answer to Problem 1PS

Solution:

The terms are $1$, $0.108$, $0.010$, $0.001$ and $0.0001$.

Explanation of Solution

Calculation:

Substitute $1$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_1& =& \frac{\left(1+1\right)}{\left(1^2+1\right)}\\ & =& \frac{2}{2}\\ & =& 1\end{array}$

Substitute $10$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{10}& =& \frac{\left(10+1\right)}{\left({10}^2+1\right)}\\ & =& \frac{11}{101}\\ & =& 0.1089\end{array}$

Substitute $100$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{100}& =& \frac{\left(100+1\right)}{\left({100}^2+1\right)}\\ & =& \frac{101}{10001}\\ & =& 0.0101\end{array}$

Substitute $1000$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{1000}& =& \frac{\left(1000+1\right)}{\left({1000}^2+1\right)}\\ & =& \frac{1001}{1000001}\\ & =& 0.0010\end{array}$

Substitute $10000$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{10000}& =& \frac{\left(10000+1\right)}{\left({10000}^2+1\right)}\\ & =& \frac{10001}{100000001}\\ & =& 0.0001\end{array}$

From the above calculation the terms are shown in following figure.

$n$	$1$	$10$	$100$	$1000$	$10000$
$a_n$	$1$	$0.1089$	$0.0101$	$0.0010$	$0.0001$

To determine

(d)

To find:

The value of $a_n$ if $n$ approaches $\infty$.

Answer to Problem 1PS

Solution:

$0$

Explanation of Solution

Calculation:

Substitute $\infty$ for $n$ in Equation (1).

$\begin{array}{rll}{a}_{\infty }& =& \frac{\left(\infty +1\right)}{\left({\infty }^2+1\right)}\\ & =& \frac{\infty }{\infty }\\ & =& 0\end{array}$

From above calculation the value of $a_n$ is $0$.

Conclusion:

Hence, the value of $a_n$ is $0$ if $n$ is $\infty$.