Chapter 11, Problem 93RE

a.

To determine

To write: The $s(n)$ as a rational function.

Answer to Problem 93RE

The correct answer is $s(n)=\frac{5n^2+9n+4}{6n^2}$ .

Explanation of Solution

Given information: In the question, $s(n)={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}-\frac{i}{n}\right)\left(\frac{1}{n}\right)}$ is given.

Calculation: Rewrite the given $s(n)$ in rational function form,

  $\begin{array}{l}s(n)={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}-\frac{i}{n}\right)\left(\frac{1}{n}\right)}\\ s(n)={\displaystyle \sum _{i=1}^n\frac{4i^2}{n^3}-}{\displaystyle \sum _{i=1}^n\frac{i}{n^2}}\\ s(n)=\frac{4}{n^3}{\displaystyle \sum _{i=1}^n{i}^2-}\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ s(n)=\frac{4}{n^3}\left\{\frac{n(n+1)(2n+1)}{6}\right\}-\frac{1}{n^2}\left\{\frac{n(n+1)}{2}\right\}\\ s(n)=\frac{5n^2+9n+4}{6n^2}\end{array}$

b.

To determine

To fill: The complete the table.

Answer to Problem 93RE

The filled correct table is:

N	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
S(n)	1	0.12	0.20	0.005	0.025

Explanation of Solution

Given information: In the question, $s(n)={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}-\frac{i}{n}\right)\left(\frac{1}{n}\right)}$ is given.

Calculation:

The filled correct table is:

N	${10}^0$	${10}^1$	${10}^2$	${10}^3$	${10}^4$
S(n)	1	0.12	0.20	0.005	0.025

c.

To determine

To find: The $\underset{n\to \infty }{\lim }s(n)$ .

Answer to Problem 93RE

The answer is $\underset{n\to \infty }{\lim }s(n)=\frac{5}{6}$ .

Explanation of Solution

Given information: In the question, $s(n)={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}-\frac{i}{n}\right)\left(\frac{1}{n}\right)}$ is given.

Calculation:

Write the expression of $s(n)$ .

  $\begin{array}{l}s(n)={\displaystyle \sum _{i=1}^n\left(\frac{4i^2}{n^2}-\frac{i}{n}\right)\left(\frac{1}{n}\right)}\\ s(n)={\displaystyle \sum _{i=1}^n\frac{4i^2}{n^3}-}{\displaystyle \sum _{i=1}^n\frac{i}{n^2}}\\ s(n)=\frac{4}{n^3}{\displaystyle \sum _{i=1}^n{i}^2-}\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}i}\\ s(n)=\frac{4}{n^3}\left\{\frac{n(n+1)(2n+1)}{6}\right\}-\frac{1}{n^2}\left\{\frac{n(n+1)}{2}\right\}\\ s(n)=\frac{5n^2+9n+4}{6n^2}\end{array}$

Apply limit on the above given function.

  $\begin{array}{c}\underset{n\to \infty }{\lim }s(n)=\frac{5n^2+9n+4}{6n^2}\\ =\frac{((5n+4)(n+1))}{(6n^2)}\\ =\frac{(n(5n+9)+4)}{(6n^2)}\\ =\frac{2}{3n^2}+\frac{3}{2n}+\frac{5}{6}\end{array}$

On substituting the limit,

  $\underset{n\to \infty }{\lim }s(n)=\frac{5}{6}$ .