Chapter 7.4, Problem 6E

a.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=\frac{n+2}{n}$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=\frac{n+2}{n}\\ x_n=\left(\frac{1+2}{1},\frac{2+2}{2},\frac{3+2}{3},\frac{4+2}{4},\frac{5+2}{5},\mathrm{....}\right)\\ =\left(\frac{3}{1},\frac{4}{2},\frac{5}{3},\frac{6}{4},\frac{7}{5},\mathrm{...}\right)\\ =\left(3,2,1.67,1.5,1.4,\mathrm{....}\right)\end{array}$

Then

  $\frac{3}{1}>\frac{4}{2}>\frac{5}{3}>\frac{6}{4}>\frac{7}{5}>\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=\frac{n+2}{n}$ is a decreasing sequence.

Therefore the sequence $x_n$ is monotone sequence.

b.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $y_n=2^{-n}$

Calculation:

Consider the sequence

  $\begin{array}{l}{y}_n=2^{-n}\\ y_n=\left(2^{-1},2^{-2},2^{-3},2^{-4},2^{-5},\mathrm{....}\right)\end{array}$

Then

  $2^{-1}>2^{-2}>2^{-3}>2^{-4}>2^{-5}>\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $y_n=2^{-n}$ is a decreasing sequence.

Therefore the sequence $y_n$ is monotone sequence.

c.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=(n-2){(n-5)}^2$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=(n-2){(}^n\\ x_n=\left((-1){(-4)}^2,(0){(-3)}^2,(1){(-2)}^2,(2){(-1)}^2,(3){(0)}^2,\mathrm{....}\right)\\ =\left(-16,0,4,2,0,16,\mathrm{...}\right)\end{array}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=(n-2){(n-5)}^2$ is both decreasing and increasing sequence.

Therefore the sequence $x_n$ is not monotone sequence.

d.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $y_n=\frac{10}{n!}$

Calculation:

Consider the sequence

  $\begin{array}{l}{y}_n=\frac{10}{n!}\\ y_n=\left(\frac{10}{1!},\frac{10}{2!},\frac{10}{3!},\frac{10}{4!},\frac{10}{5!},\mathrm{....}\right)\end{array}$

Then

  $\frac{10}{1!}>\frac{10}{2!}>\frac{10}{3!}>\frac{10}{4!}>\frac{10}{5!}>\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $y_n=\frac{10}{n!}$ is a decreasing sequence.

Therefore the sequence $y_n$ is monotone sequence.

e.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=2^{-n}\left(\frac{n+2}{n}\right)$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=2^{-n}\left(\frac{n+2}{n}\right)\\ x_n=\left(2^{-1}\left(\frac{1+2}{1}\right),2^{-2}\left(\frac{2+2}{2}\right),2^{-3}\left(\frac{3+2}{3}\right),2^{-4}\left(\frac{4+2}{4}\right),2^{-5}\left(\frac{5+2}{5}\right),\mathrm{....}\right)\end{array}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=2^{-n}\left(\frac{n+2}{n}\right)$ is a decreasing sequence.

Therefore the sequence $x_n$ is monotone sequence.

f.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=\frac{2n-5}{n+3}$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=\frac{2n-5}{n+3}\\ x_{n+1}=\frac{2(n+1)-5}{n+4}\\ =\frac{2n-3}{n+4}\end{array}$

Then

  $x_{n+1}>x_n$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=\frac{2n-5}{n+3}$ is anincreasing sequence.

Therefore the sequence $x_n$ is monotone sequence.

g.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $y_n=\frac{n!}{n^n}$

Calculation:

Consider the sequence

  $\begin{array}{l}{y}_n=\frac{n!}{n^n}\\ y_n=\left(\frac{1!}{1},\frac{2!}{2^2},\frac{3!}{3^3},\frac{4!}{4^4},\frac{5!}{5^5},\mathrm{....}\right)\end{array}$

Then

  $\frac{1!}{1}>\frac{2!}{2^2}>\frac{3!}{3^3}>\frac{4!}{4^4}>\frac{5!}{5^5}>\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $y_n=\frac{n!}{n^n}$ is a decreasing sequence.

Therefore the sequence $y_n$ is monotone sequence.

h.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=\frac{n!}{n+1}$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=\frac{n!}{n+1}\\ x_n=\left(\frac{1!}{2},\frac{2!}{3},\frac{3!}{4},\frac{4!}{5},\frac{5!}{6},\mathrm{....}\right)\end{array}$

Then

  $\frac{1!}{2}<\frac{2!}{3}<\frac{3!}{4}<\frac{4!}{5}<\frac{5!}{6}<\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=\frac{n!}{n+1}$ is an increasing sequence.

Therefore the sequence $x_n$ is monotone sequence.

i.

To determine

To prove whether the sequence is monotone or not.

Explanation of Solution

Given:

  $x_n=\sqrt{n+1}$

Calculation:

Consider the sequence

  $\begin{array}{l}{x}_n=\sqrt{n+1}\\ x_n=\left(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6},\mathrm{....}\right)\end{array}$

Then

  $\sqrt{2}<\sqrt{3}<\sqrt{4}<\sqrt{5}<\sqrt{6}<\mathrm{..}$

The monotonic sequence should either be increasing or decreasing.

Since, the given sequence $x_n=\sqrt{n+1}$ is anincreasing sequence.

Therefore the sequence $x_n$ is monotone sequence.