Chapter 12.5, Problem 16E

a.

To determine

ToDetermineRewrite the sum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$ as rational function $S\left(n\right)$

Answer to Problem 16E

The sum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$ as a rational function is $S\left(n\right)=1+\frac{4}{n}$

Explanation of Solution

Given:

The sum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$

Concept Used:

  ${\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}$

Calculation:

Given the sum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$

i.e., $\begin{array}{c}S={\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}\\ =\frac{1}{n^2}{\displaystyle \sum _{i=1}^n\left(2i+3\right)}\\ =\frac{1}{n^2}\left(2{\displaystyle \sum _{i=1}^n\left(i\right)}+{\displaystyle \sum _{i=1}^n\left(3\right)}\right){\displaystyle \sum _{i=1}^{n}i}=\frac{n\left(n+1\right)}{2}\end{array}$

  $\begin{array}{c}=\frac{1}{n^2}\left(2\left(\frac{n\left(n+1\right)}{2}\right)+3\left(n\right)\right)\\ =\frac{1}{n^2}\left(n\left(n+1\right)+3n\right)\\ =\frac{1}{n^2}\left(n^2+n+3n\right)\\ =\frac{1}{n^2}\left(n^2+4n\right)\end{array}$

  $\begin{array}{c}=\frac{n^2}{n^2}+\frac{4n}{n^2}\\ =1+\frac{4n}{n^2}\\ =1+\frac{4}{n}\end{array}$

Therefore, the sum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$ as a rational function is $S\left(n\right)=1+\frac{4}{n}$

Conclusion:

Thesum ${\displaystyle \sum _{i=1}^n\frac{2i+3}{n^2}}$ as a rational function is $S\left(n\right)=1+\frac{4}{n}$

b.

To determine

Use the $S\left(n\right)$ found in the previous part of the question to complete the below table.

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$

Answer to Problem 16E

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$5$	$1.4$	$1.04$	$1.004$	$1.0004$

Explanation of Solution

Given:

The $S\left(n\right)=1+\frac{4}{n}$

Concept Used:

  $a^0=1$

Calculation:

Given $S\left(n\right)=1+\frac{4}{n}$

For $n={10}^0$

  $\begin{array}{c}S\left({10}^0\right)=1+\frac{4}{{10}^0}\\ =1+\frac{4}{1}{a}^0=1\\ =1+5\\ =5\end{array}$

For $n={10}^1$

  $\begin{array}{c}S\left({10}^1\right)=1+\frac{4}{{10}^1}\\ =1+\frac{4}{10}\\ =1+0.4\\ =1.4\end{array}$

For $n={10}^2$

  $\begin{array}{c}S\left({10}^2\right)=1+\frac{4}{{10}^2}\\ =1+\frac{4}{100}\\ =1+0.04\\ =1.04\end{array}$

For $n={10}^3$

  $\begin{array}{c}S\left({10}^3\right)=1+\frac{4}{{10}^3}\\ =1+\frac{4}{1000}\\ =1+0.004\\ =1.004\end{array}$

For $n={10}^4$

  $\begin{array}{c}S\left({10}^4\right)=1+\frac{4}{{10}^4}\\ =1+\frac{4}{10000}\\ =1+0.0004\\ =1.0004\end{array}$

Completing the table

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$5$	$1.4$	$1.04$	$1.004$	$1.0004$

Conclusion:

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$5$	$1.4$	$1.04$	$1.004$	$1.0004$

c.

To determine

The $\underset{n\to \infty }{\lim }S\left(n\right)$ , for $S\left(n\right)=1+\frac{4}{n}$ found in part a.

Answer to Problem 16E

  $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}1$

Explanation of Solution

Given:

The $S\left(n\right)=1+\frac{4}{n}$

Concept Used:

  $\underset{n\to c}{\lim }\left(a+b\right)=\underset{n\to c}{\lim }a+\underset{n\to c}{\lim }b$

Calculation:

For $S\left(n\right)=1+\frac{4}{n}$

  $\begin{array}{c}\underset{n\to \infty }{\lim }S\left(n\right)=\underset{n\to \infty }{\lim }\left(1+\frac{4}{n}\right)\\ =\left(\underset{n\to \infty }{\lim }1+\underset{n\to \infty }{\lim }\frac{4}{n}\right)\underset{n\to c}{\lim }\left(a+b\right)=\underset{n\to c}{\lim }a+\underset{n\to c}{\lim }b\end{array}$

  $\begin{array}{c}=\left(1+0\right)\text{ Substituting limit}\\ =1\end{array}$

Therefore, $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}1$

Conclusion:

  $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}1$