Chapter 8.1, Problem 50E

To determine

The sequence is increasing, decreasing or not monotonic and bounded or not bounded.

Answer to Problem 50E

The sequence is monotone increasing and the sequence is bounded.

Explanation of Solution

Given information: The sequence is $a_n=\frac{2n-3}{3n+4}$

Definitions:

Monotone increasing: A sequence ${\left\{x_n\right\}}_n$ is said to be monotone increasing, iff $\forall n,x_{n+1}\ge x_n$

Monotone decreasing: A sequence ${\left\{x_n\right\}}_n$ is said to be monotone decreasing iff $\forall n,x_{n+1}\le x_n.$

Given sequence,. $a_n=\frac{2n-3}{3n+4}$

Now we find $a_{n+1}$

  $a_{n+1}=\frac{2n-1}{3n+7}$

From the above definition, we have $a_{n+1}$

  $-a_n$

  $=\frac{2n-1}{3n+7}-\frac{2n-3}{3n+4}$

  $=\frac{(2n-1)(3n+4)-(2n-3)(3n+7)}{(3n+7)(3n+4)}$

  $\begin{array}{l}=\frac{6n^2-3n+8n-4-6n^2-14n+9n+21}{(3n+7)(3n+4)}\\ =\frac{17}{(3n+7)(3n+4)}\end{array}$

Thus, $\frac{17}{(3n+7)(3n+4)}>0$

Hence, $a_{n+1}$ $>a_n$

From the above definition , the sequence is monotone increasing .

Definition of a bounded sequence: A sequence ${\left\{x_n\right\}}_n$ is said to be bounded if $\exists$ a real positive number K s.t. $\forall n\in \mathbb{N},$

  $\left|x_n\right|\le K\text{ i}\text{.e }-K\le x_n\le -K.$

Now let,

  $\frac{2n-3}{3n+4}\le \frac{2}{3}$

To justify that the guess is correct.

In order that this inequality may be true we should have

  $6n-9\le 6n+8,\forall n$ and $6n-9\le 6n+8\Rightarrow -9\le 8$ which is true , whatever n we may take

Hence, the given sequence is bounded.