Chapter 12.5, Problem 17E

a.

To determine

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

Answer to Problem 17E

The sum ${\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}+\frac{2}{n}\right)\left(\frac{1}{n}\right)}$ as a rational function is $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

Explanation of Solution

Given:

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

Concept Used:

  ${\displaystyle \sum _{i=1}^n{i}^2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

Calculation:

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

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

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

  $=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

Therefore, the sum ${\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}+\frac{2}{n}\right)\left(\frac{1}{n}\right)}$ as a rational function is $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

Conclusion:

Thesum ${\displaystyle \sum _{i=1}^n\left(\frac{i^2}{n^3}+\frac{2}{n}\right)\left(\frac{1}{n}\right)}$ as a rational function is $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

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 17E

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$3$	$0.02385$	$0.000233835$	$2.3338335\times {10}^{-6}$	$2.333383335\times {10}^{-8}$

Explanation of Solution

Given:

The $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

Concept Used:

  $a^0=1$

Calculation:

Given $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

For $n={10}^0$

  $\begin{array}{c}S\left({10}^0\right)=\frac{1}{3{\left({10}^0\right)}^2}+\frac{1}{2{\left({10}^0\right)}^3}+\frac{1}{6{\left({10}^0\right)}^4}+\frac{2}{{\left({10}^0\right)}^2}\\ =\frac{1}{3\left(1\right)}+\frac{1}{2\left(1\right)}+\frac{1}{6\left(1\right)}+\frac{2}{\left(1\right)}{a}^0=1\\ =\frac{1}{3}+\frac{1}{2}+\frac{1}{6}+2\\ =3\end{array}$

For $n={10}^1$

  $\begin{array}{c}S\left({10}^1\right)=\frac{1}{3{\left({10}^1\right)}^2}+\frac{1}{2{\left({10}^1\right)}^3}+\frac{1}{6{\left({10}^1\right)}^4}+\frac{2}{{\left({10}^1\right)}^2}\\ =\frac{1}{3\left({10}^2\right)}+\frac{1}{2\left({10}^3\right)}+\frac{1}{6\left({10}^4\right)}+\frac{2}{\left({10}^2\right)}\\ =0.02385\end{array}$

For $n={10}^2$

  $\begin{array}{c}S\left({10}^2\right)=\frac{1}{3{\left({10}^2\right)}^2}+\frac{1}{2{\left({10}^2\right)}^3}+\frac{1}{6{\left({10}^2\right)}^4}+\frac{2}{{\left({10}^2\right)}^2}\\ =\frac{1}{3\left({10}^4\right)}+\frac{1}{2\left({10}^6\right)}+\frac{1}{6\left({10}^8\right)}+\frac{2}{\left({10}^4\right)}\\ =0.000233835\end{array}$

For $n={10}^3$

  $\begin{array}{c}S\left({10}^3\right)=\frac{1}{3{\left({10}^3\right)}^2}+\frac{1}{2{\left({10}^3\right)}^3}+\frac{1}{6{\left({10}^3\right)}^4}+\frac{2}{{\left({10}^3\right)}^2}\\ =\frac{1}{3\left({10}^6\right)}+\frac{1}{2\left({10}^9\right)}+\frac{1}{6\left({10}^{12}\right)}+\frac{2}{\left({10}^6\right)}\\ =2.3338335\times {10}^{-6}\end{array}$

For $n={10}^4$

  $\begin{array}{c}S\left({10}^4\right)=\frac{1}{3{\left({10}^4\right)}^2}+\frac{1}{2{\left({10}^4\right)}^3}+\frac{1}{6{\left({10}^4\right)}^4}+\frac{2}{{\left({10}^4\right)}^2}\\ =\frac{1}{3\left({10}^8\right)}+\frac{1}{2\left({10}^{12}\right)}+\frac{1}{6\left({10}^{16}\right)}+\frac{2}{\left({10}^8\right)}\\ =2.333383335\times {10}^{-8}\end{array}$

Completing the table

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$3$	$0.02385$	$0.000233835$	$2.3338335\times {10}^{-6}$	$2.333383335\times {10}^{-8}$

Conclusion:

The completed table is

$n$	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
$S\left(n\right)$	$3$	$0.02385$	$0.000233835$	$2.3338335\times {10}^{-6}$	$2.333383335\times {10}^{-8}$

c.

To determine

The $\underset{n\to \infty }{\lim }S\left(n\right)$ , for $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$ found in part a.

Answer to Problem 17E

  $\underset{n\to \infty }{\lim }S\left(n\right)=0$

Explanation of Solution

Given:

The $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

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 $S\left(n\right)=\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}$

  $\begin{array}{c}\underset{n\to \infty }{\lim }S\left(n\right)=\underset{n\to \infty }{\lim }\left(\frac{1}{3n^2}+\frac{1}{2n^3}+\frac{1}{6n^4}+\frac{2}{n^2}\right)\\ =\underset{n\to \infty }{\lim }\left(\frac{1}{3n^2}\right)+\underset{n\to \infty }{\lim }\left(\frac{1}{2n^3}\right)+\underset{n\to \infty }{\lim }\left(\frac{1}{6n^4}\right)+\underset{n\to \infty }{\lim }\left(\frac{2}{n^2}\right)\underset{n\to c}{\lim }\left(a+b\right)=\underset{n\to c}{\lim }a+\underset{n\to c}{\lim }b\\ =0+0+0+0\\ =0\end{array}$

Therefore, $\underset{n\to \infty }{\lim }S\left(n\right)=0$

Conclusion:

  $\underset{n\to \infty }{\lim }S\left(n\right)=0$