Chapter 12.5, Problem 13E

a.

To determine

ToDetermineRewrite the sum ${\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}$ as rational function $S\left(n\right)$

Answer to Problem 13E

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

Explanation of Solution

Given:

The sum ${\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}$

Concept Used:

  ${\displaystyle \sum _{i=1}^n{i}^3}=\frac{n^2{\left(n+1\right)}^2}{4}$

  $\frac{a^m}{a^n}=a^{m-n}$

Calculation:

Given the sum ${\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}$

i.e., $\begin{array}{c}S={\displaystyle \sum _{i=1}^n\frac{i^3}{n^4}}\\ =\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\\ =\frac{1}{n^4}{\displaystyle \sum _{i=1}^n{i}^3}\\ =\frac{1}{n^4}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right){\displaystyle \sum _{i=1}^n{i}^3}=\frac{n^2{\left(n+1\right)}^2}{4}\end{array}$

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

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

Conclusion:

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

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 13E

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$1$	$0.3025$	$0.255025$	$0.25050025$	$0.2500500025$

Explanation of Solution

Given:

The $S\left(n\right)=\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2$

Concept Used:

  $a^0=1$

Calculation:

Given $S\left(n\right)=\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2$

For $n={10}^0$

  $\begin{array}{c}S\left({10}^0\right)=\frac{1}{4}{\left(\frac{{10}^0+1}{{10}^0}\right)}^2\\ =\frac{1}{4}{\left(\frac{1+1}{1}\right)}^2{a}^0=1\\ =\frac{1}{4}{\left(\frac{2}{1}\right)}^2\\ =\frac{1}{4}{\left(2\right)}^2\end{array}$

  $\begin{array}{c}=\frac{4}{4}\\ =1\end{array}$

For $n={10}^1$

  $\begin{array}{c}S\left({10}^1\right)=\frac{1}{4}{\left(\frac{{10}^1+1}{{10}^1}\right)}^2\\ =\frac{1}{4}{\left(\frac{10+1}{10}\right)}^2\\ =\frac{1}{4}{\left(\frac{11}{10}\right)}^2\\ =\frac{1}{4}{\left(1.1\right)}^2\end{array}$

  $\begin{array}{c}=\frac{1}{4}\left(1.21\right)\\ =0.3025\end{array}$

For $n={10}^2$

  $\begin{array}{c}S\left({10}^2\right)=\frac{1}{4}{\left(\frac{{10}^2+1}{{10}^2}\right)}^2\\ =\frac{1}{4}{\left(\frac{100+1}{100}\right)}^2\\ =\frac{1}{4}{\left(\frac{101}{100}\right)}^2\\ =\frac{1}{4}{\left(1.01\right)}^2\end{array}$

  $\begin{array}{c}=\frac{1}{4}\left(1.0201\right)\\ =0.255025\end{array}$

For $n={10}^3$

  $\begin{array}{c}S\left({10}^3\right)=\frac{1}{4}{\left(\frac{{10}^3+1}{{10}^3}\right)}^2\\ =\frac{1}{4}{\left(\frac{1000+1}{1000}\right)}^2\\ =\frac{1}{4}{\left(\frac{1001}{1000}\right)}^2\\ =\frac{1}{4}{\left(1.001\right)}^2\end{array}$

  $\begin{array}{c}=\frac{1}{4}\left(1.002001\right)\\ =0.25050025\end{array}$

  $\begin{array}{c}S\left({10}^4\right)=\frac{1}{4}{\left(\frac{{10}^4+1}{{10}^4}\right)}^2\\ =\frac{1}{4}{\left(\frac{10000+1}{10000}\right)}^2\\ =\frac{1}{4}{\left(\frac{10001}{10000}\right)}^2\\ =\frac{1}{4}{\left(1.0001\right)}^2\end{array}$

  $\begin{array}{c}=\frac{1}{4}\left(1.00020001\right)\\ =0.2500500025\end{array}$

Completing the table

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$1$	$0.3025$	$0.255025$	$0.25050025$	$0.2500500025$

Conclusion:

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$1$	$0.3025$	$0.255025$	$0.25050025$	$0.2500500025$

c.

To determine

The $\underset{n\to \infty }{\lim }S\left(n\right)$ , for $S\left(n\right)=\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2$ found in part a.

Answer to Problem 13E

  $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}\frac{1}{4}$

Explanation of Solution

Given:

The $S\left(n\right)=\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2$

Concept Used:

  $\underset{n\to c}{\lim }{\left(a_n\right)}^k={\left(\underset{n\to c}{\lim }{a}_n\right)}^k$

Calculation:

For $S\left(n\right)=\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2$

  $\begin{array}{c}\underset{n\to \infty }{\lim }S\left(n\right)=\underset{n\to \infty }{\lim }\frac{1}{4}{\left(\frac{n+1}{n}\right)}^2\\ =\frac{1}{4}\underset{n\to \infty }{\lim }{\left(\frac{n+1}{n}\right)}^2\\ =\frac{1}{4}\underset{n\to \infty }{\lim }{\left(\frac{n}{n}+\frac{1}{n}\right)}^2\\ =\frac{1}{4}\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^2\end{array}$

  $\begin{array}{c}=\frac{1}{4}{\left(\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\right)}^2\underset{n\to c}{\lim }{\left(a_n\right)}^k={\left(\underset{n\to c}{\lim }{a}_n\right)}^k\\ =\frac{1}{4}{\left(1+0\right)}^2\text{ Substituting limit}\\ \text{=}\frac{1}{4}\end{array}$

Therefore, $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}\frac{1}{4}$

Conclusion:

  $\underset{n\to \infty }{\lim }S\left(n\right)\text{=}\frac{1}{4}$