Chapter 8.2, Problem 9E

To determine

The sequence is convergent and the series is convergent.

Answer to Problem 9E

The sequence is convergent. the series is convergent.

Explanation of Solution

Given information: The sequence is $a_n=\frac{2n}{3n+1}$

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,. $a_n=\frac{2n}{3n+1}$

Now we find $a_{n+1}$

  $a_{n+1}=\frac{2n+2}{3n+4}$

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

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

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

  $\begin{array}{l}=\frac{6n^2+2n+6n+2-6n^2-8n}{(3n+4)(3n+1)}\\ =\frac{2}{(3n+4(3n+1))}\end{array}$

Thus, $\frac{2}{(3n+1)(3n+4)}>0$

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,

  $\frac{2n}{3n+1}\le \frac{2}{3}$

To justify that the guess is correct.

In order that this inequality may be true we should have

  $6n\le 6n+2,\forall n$ and $6n\le 6n+2\Rightarrow 0\le 2$ which is true , whatever n we may take

Hence, the given sequence is bounded.

Since, the sequence is bounded and monotone increasing

then the sequence is convergent.

Given, $a_n=\frac{2}{3+\frac{1}{n}}$

Take limit both side, we have

  $\begin{array}{l}\underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{2}{3+\frac{1}{n}}\\ \underset{x\to \infty }{\lim }{a}_n=\frac{2}{3}\end{array}$

So the series is convergent.