Chapter 10, Problem 1GYR

To determine

Provide notes on an infinite sequence, the meaning of the convergent sequence, and the meaning of divergent sequence with example.

Explanation of Solution

The infinite sequence of numbers is a function whose domain is the set of positive integers.

Example:

Consider the following sequence is $2,4,6,8,\cdots ,2n$

Provide the sequence of each term as follows.

$\begin{array}{c}{a}_1=2\\ a_2=4\\ a_3=6\\ a_4=8\\ a_n=2n\end{array}$

The formula for $n^{\text{th}}$ term of the sequence is $a_n=2n$, where $n\ge 1$.

Hence, the formula for $n^{\text{th}}$ term of the sequence is $\underset{\_}{a_n=2n,n\ge 1}$.

Provide the sequence to be converge and diverge as shown below.

The sequence $\left\{a_n\right\}$ converges to the number L if for every positive number $\epsilon$ there corresponds an integer N such that

$\left|a_n-L\right|<\epsilon$ whenever $n>N$.

If no such number L exists, we say that $\left\{a_n\right\}$ diverges.

If $\left\{a_n\right\}$ converges to L, we write ${\lim }_{n\to \infty }{a}_n=L$ or simply $a_n=L$, and call L the limit of the sequence.

Example:

For converge sequence:

Consider the nth formula for the sequence function is $a_n=2+{\left(0.1\right)}^n$.

Find the limit of sequence as follows:

$\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(2+{\left(0.1\right)}^n\right)\\ =\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }{\left(0.1\right)}^n\\ =\underset{n\to \infty }{\lim }2+\underset{n\to \infty }{\lim }\frac{1}{{10}^n}\\ =2+\frac{1}{{10}^{\infty }}\\ =2+0\\ =2\end{array}$

The limit of the sequence is 2.

Hence, the sequence is converge.

For diverge sequence:

The nth formula for the sequence function is $a_n=\frac{2n+1}{1-3\sqrt{n}}$.

Find the limit of sequence as follows:

$\begin{array}{c}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\left(\frac{2n+1}{1-3\sqrt{n}}\right)\\ =\underset{n\to \infty }{\lim }\frac{\sqrt{n}\left(2\sqrt{n}+\frac{1}{\sqrt{n}}\right)}{\sqrt{n}\left(\frac{1}{\sqrt{n}}-3\right)}\\ =\underset{n\to \infty }{\lim }\frac{2\sqrt{n}+\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}-3}\\ =\frac{\underset{n\to \infty }{\lim }2\sqrt{n}+\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}}{\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n}}-\underset{n\to \infty }{\lim }3}\end{array}$

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

The limit of the sequence does not exists.

Therefore, the sequence is diverge.