Chapter 11.5, Problem 16E

(a)

To determine

To find: The rational function $S\left(n\right)$ for sum.

Answer to Problem 16E

The rational function $S\left(n\right)$ for sum is equal to $1+\frac{7}{n}+\frac{1}{n^2}$ .

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}$ .

Calculation:

Simplify the given sum.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)}\\ =\frac{1}{n}\left({\displaystyle \sum _{i=1}^n\frac{5}{n}}+{\displaystyle \sum _{i=1}^n\frac{4i^3}{n^3}}\right)\\ =\frac{1}{n}\left(\frac{5}{n}{\displaystyle \sum _{i=1}^{n}1}+\frac{4}{n^3}{\displaystyle \sum _{i=1}^n{i}^3}\right)\end{array}$

Substitute $n$ for the sum ${\displaystyle \sum _{i=1}^{n}1}$ and $\frac{n^2{\left(n+1\right)}^2}{4}$ for ${\displaystyle \sum _{i=1}^n{i}^3}$ in the sum.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}=\frac{1}{n}\left(\frac{5}{n}{\displaystyle \sum _{i=1}^{n}1}+\frac{4}{n^3}{\displaystyle \sum _{i=1}^n{i}^3}\right)\\ =\frac{1}{n}\left(\frac{5}{n}\left(n\right)+\frac{4}{n^3}\left(\frac{n^2{\left(n+1\right)}^2}{4}\right)\right)\\ =\frac{1}{n}\left(5+\frac{{\left(n+1\right)}^2}{n}\right)\\ =\frac{1}{n}\left(5+\frac{n^2+1+2n}{n}\right)\end{array}$

Further simplify,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}=\frac{1}{n}\left(5+\frac{n^2+1+2n}{n}\right)\\ =\frac{1}{n}\left(\frac{5n+n^2+1+2n}{n}\right)\\ =\frac{1}{n}\left(\frac{n^2+7n+1}{n}\right)\\ =1+\frac{7}{n}+\frac{1}{n^2}\end{array}$

Therefore, the rational function $S\left(n\right)$ for sum is equal to $1+\frac{7}{n}+\frac{1}{n^2}$ .

(b)

To determine

To find: The complete table given below with the help of rational function $S\left(n\right)$ :

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

Answer to Problem 16E

The complete table is given below:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$9$	$1.71$	$1.0701$	$1.007$	$1.0007$

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}$ .

Calculation:

Substitute ${10}^0$ for $n$ in the function for $S\left(n\right)$ .

  $\begin{array}{c}S\left(n\right)=1+\frac{7}{n}+\frac{1}{n^2}\\ S\left({10}^0\right)=1+\frac{7}{{10}^0}+\frac{1}{{\left({10}^0\right)}^2}\\ S\left({10}^0\right)=1+\frac{7}{1}+\frac{1}{1}\\ S\left({10}^0\right)=1+7+1\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^0\right)=1+7+1\\ S\left({10}^0\right)=9\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^0$ is equal to $9$ .

Substitute ${10}^1$ for $n$ in the function for $S\left(n\right)$ .

  $\begin{array}{c}S\left(n\right)=1+\frac{7}{n}+\frac{1}{n^2}\\ S\left({10}^1\right)=1+\frac{7}{{10}^1}+\frac{1}{{\left({10}^1\right)}^2}\\ S\left({10}^1\right)=1+\frac{7}{10}+\frac{1}{100}\\ S\left({10}^1\right)=1+0.7+0.01\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^1\right)=1+0.7+0.01\\ S\left({10}^1\right)=1.71\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^1$ is equal to $1.71$ .

Substitute ${10}^2$ for $n$ in the function for $S\left(n\right)$ .

  $\begin{array}{c}S\left(n\right)=1+\frac{7}{n}+\frac{1}{n^2}\\ S\left({10}^2\right)=1+\frac{7}{{10}^2}+\frac{1}{{\left({10}^2\right)}^2}\\ S\left({10}^2\right)=1+\frac{7}{100}+\frac{1}{10000}\\ S\left({10}^2\right)=1+0.07+0.0001\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^2\right)=1+0.07+0.0001\\ S\left({10}^2\right)=1.0701\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^2$ is equal to $1.0701$ .

Substitute ${10}^3$ for $n$ in the function for $S\left(n\right)$ .

  $\begin{array}{c}S\left(n\right)=1+\frac{7}{n}+\frac{1}{n^2}\\ S\left({10}^3\right)=1+\frac{7}{{10}^3}+\frac{1}{{\left({10}^3\right)}^2}\\ S\left({10}^3\right)=1+\frac{7}{1000}+\frac{1}{1000000}\\ S\left({10}^3\right)=1+0.007+0.000001\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^3\right)=1+0.007+0.000001\\ S\left({10}^3\right)\approx 1.007\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^3$ is equal to $1.007$ .

Substitute ${10}^4$ for $n$ in the function for $S\left(n\right)$ .

  $\begin{array}{c}S\left(n\right)=1+\frac{7}{n}+\frac{1}{n^2}\\ S\left({10}^4\right)=1+\frac{7}{{10}^4}+\frac{1}{{\left({10}^4\right)}^2}\\ S\left({10}^4\right)=1+\frac{7}{10000}+\frac{1}{100000000}\\ S\left({10}^4\right)=1+0.0007+0.00000001\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^4\right)=1+0.0007+0.00000001\\ S\left({10}^4\right)\approx 1.0007\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^4$ is equal to $1.0007$ .

Therefore, the complete table is given below:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$9$	$1.71$	$1.0701$	$1.007$	$1.0007$

(c)

To determine

To find: The value of $\underset{n\to \infty }{\lim }S\left(n\right)$ .

Answer to Problem 16E

The value of $\underset{n\to \infty }{\lim }S\left(n\right)$ is equal to $1$ .

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\left(\frac{5}{n}+\frac{4i^3}{n^3}\right)\left(\frac{1}{n}\right)}$ .

Calculation:

Simplify the limit.

  $\begin{array}{c}\underset{n\to \infty }{\lim }S\left(n\right)=\underset{n\to \infty }{\lim }\left(1+\frac{7}{n}+\frac{1}{n^2}\right)\\ =\underset{n\to \infty }{\lim }\left(1\right)+\underset{n\to \infty }{\lim }\left(\frac{7}{n}\right)+\underset{n\to \infty }{\lim }\left(\frac{1}{n^2}\right)\\ =1+0+0\\ =1\end{array}$

Therefore, the value of $\underset{n\to \infty }{\lim }S\left(n\right)$ is equal to $1$ .