Chapter 9, Problem 8RE

To determine

To determine the convergence or divergence of the sequence with given $nth$ term. Also to determine limit if the sequence is convergent

Answer to Problem 8RE

The given sequence diverges.

Explanation of Solution

Given information :

  $a_n={\left(-1\right)}^n\frac{3n-1}{n+2}$

Concept used:

  $\underset{n\to \infty }{\lim }{a}_n=l$ where $l$ is limit then the sequence is said to converge and if limit does not exist then sequence is said to be divergent.

Given $a_n={\left(-1\right)}^n\frac{3n-1}{n+2}$

Taking $\underset{n\to \infty }{\lim }$ on both sides

  $\begin{array}{l}\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\times \frac{3n-1}{n+2}\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\times \underset{n\to \infty }{\lim }\frac{n\left(3-\frac{1}{n}\right)}{n\left(1+\frac{1}{n}\right)}\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\times \left(\frac{3+\frac{1}{\infty }}{1+\frac{1}{\infty }}\right)\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\times \left(\frac{3+0}{1+0}\right)\\ \underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{\left(-1\right)}^n\times 3\\ \underset{n\to \infty }{\lim }{a}_n=3\times \underset{n\to \infty }{\lim }{\left(-1\right)}^n\end{array}$

  $\begin{array}{l}Ifn\text{ is even }then{a}_n=3\\ Ifn\text{ is odd }then{a}_n=-3\end{array}$

This sequence diverges. In fact, the terms with n even sequence approaches 3 while the terms with n odd approaches $-1$