Chapter 11.5, Problem 13E

(a)

To determine

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

Answer to Problem 13E

The rational function $S\left(n\right)$ for sum is equal to $\frac{3}{2}n+3+\frac{3}{2n}+\frac{5}{n^2}$ .

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}$ .

Calculation:

Simplify the given sum.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}=\frac{1}{n^3}{\displaystyle \sum _{i=1}^n\left(5+6i^2\right)}\\ =\frac{1}{n^3}\left({\displaystyle \sum _{i=1}^{n}5}+{\displaystyle \sum _{i=1}^n\left(6i^2\right)}\right)\\ =\frac{1}{n^3}\left(5{\displaystyle \sum _{i=1}^{n}1}+6{\displaystyle \sum _{i=1}^n{i}^2}\right)\end{array}$

Substitute $n$ for the sum ${\displaystyle \sum _{i=1}^{n}1}$ and $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ for ${\displaystyle \sum _{i=1}^n{i}^3}$ in the sum.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}=\frac{1}{n^3}\left(5{\displaystyle \sum _{i=1}^{n}1}+6{\displaystyle \sum _{i=1}^n{i}^2}\right)\\ =\frac{1}{n^3}\left(5\left(n\right)+6\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right)\right)\\ =\frac{1}{n^3}\left(5n+\left(n^2+n\right)\left(2n+1\right)\right)\\ =\frac{1}{n^3}\left(5n+\left(2n^3+n^2+2n^2+n\right)\right)\end{array}$

Further simplify,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}=\frac{1}{n^3}\left(5n+\left(2n^3+n^2+2n^2+n\right)\right)\\ =\frac{1}{n^3}\left(2n^3+3n^2+6n\right)\\ =\frac{2n^3}{n^3}+\frac{3n^2}{n^3}+\frac{6n}{n^3}\\ =2+\frac{3}{n}+\frac{6}{n^2}\end{array}$

Therefore, the rational function $S\left(n\right)$ for sum is equal to $2+\frac{3}{n}+\frac{6}{n^2}$ .

(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 13E

The complete table is given below:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$11$	$2.36$	$2.0306$	$2.003$	$2.0003$

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}$ .

Calculation:

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

  $\begin{array}{c}S\left(n\right)=2+\frac{3}{n}+\frac{6}{n^2}\\ S\left({10}^0\right)=2+\frac{3}{{10}^0}+\frac{6}{{\left({10}^0\right)}^2}\\ S\left({10}^0\right)=2+\frac{3}{1}+\frac{6}{1}\\ S\left({10}^0\right)=2+3+6\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^0\right)=2+3+6\\ S\left({10}^0\right)=11\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^0$ is equal to $11$ .

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

  $\begin{array}{c}S\left(n\right)=2+\frac{3}{n}+\frac{6}{n^2}\\ S\left({10}^1\right)=2+\frac{3}{{10}^1}+\frac{6}{{\left({10}^1\right)}^2}\\ S\left({10}^1\right)=2+\frac{3}{10}+\frac{6}{100}\\ S\left({10}^1\right)=2+0.3+0.06\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^1\right)=2+0.3+0.06\\ S\left({10}^1\right)=2.36\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^1$ is equal to $2.36$ .

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

  $\begin{array}{c}S\left(n\right)=2+\frac{3}{n}+\frac{6}{n^2}\\ S\left({10}^2\right)=2+\frac{3}{{10}^2}+\frac{6}{{\left({10}^2\right)}^2}\\ S\left({10}^2\right)=2+\frac{3}{100}+\frac{6}{10000}\\ S\left({10}^2\right)=2+0.03+0.0006\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^2\right)=2+0.03+0.0006\\ S\left({10}^2\right)=2.0306\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^2$ is equal to $2.0306$ .

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

  $\begin{array}{c}S\left(n\right)=2+\frac{3}{n}+\frac{6}{n^2}\\ S\left({10}^3\right)=2+\frac{3}{{10}^3}+\frac{6}{{\left({10}^3\right)}^2}\\ S\left({10}^3\right)=2+\frac{3}{1000}+\frac{6}{1000000}\\ S\left({10}^3\right)=2+0.003+0.000006\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^3\right)=2+0.003+0.000006\\ S\left({10}^3\right)\approx 2.003\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^3$ is equal to $2.003$ .

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

  $\begin{array}{c}S\left(n\right)=2+\frac{3}{n}+\frac{6}{n^2}\\ S\left({10}^4\right)=2+\frac{3}{{10}^4}+\frac{6}{{\left({10}^4\right)}^2}\\ S\left({10}^4\right)=2+\frac{3}{10000}+\frac{6}{100000000}\\ S\left({10}^4\right)=2+0.0003+0.00000006\end{array}$

Further simplify,

  $\begin{array}{c}S\left({10}^4\right)=2+0.0003+0.00000006\\ S\left({10}^4\right)\approx 2.0003\end{array}$

So, the value of $S\left(n\right)$ for $n={10}^4$ is equal to $2.0003$ .

Therefore, the complete table is given below:

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$11$	$2.36$	$2.0306$	$2.003$	$2.0003$

(c)

To determine

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

Answer to Problem 13E

The value of $\underset{n\to \infty }{\lim }S\left(n\right)$ is equal to $2$ .

Explanation of Solution

Given information:

The sum is ${\displaystyle \sum _{i=1}^n\frac{5+6i^2}{n^3}}$ .

Calculation:

Simplify the limit.

  $\begin{array}{c}\underset{n\to \infty }{\lim }S\left(n\right)=\underset{n\to \infty }{\lim }\left(2+\frac{3}{n}+\frac{6}{n^2}\right)\\ =\underset{n\to \infty }{\lim }\left(2\right)+\underset{n\to \infty }{\lim }\left(\frac{3}{n}\right)+\underset{n\to \infty }{\lim }\left(\frac{6}{n^2}\right)\\ =2+0+0\\ =2\end{array}$

Therefore, the value of $\underset{n\to \infty }{\lim }S\left(n\right)$ is equal to $2$ .