Chapter 14.2, Problem 36E

To determine

To find: The series is convergent or divergent by use of ratio test.

Answer to Problem 36E

The given series is convergent.

Explanation of Solution

Given:

The given series is as:

  $\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\cdots \frac{n}{3^n}+\cdots$

The $\text{}{a}_n$ is given as:

  $a_n=\frac{n}{3^n}$

The ratio test is given as:

  $r=\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$

The convergent or divergent is check as:

  $\begin{array}{c}r=\underset{n\to \infty }{\lim }\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^n}}\\ =\underset{n\to \infty }{\lim }\frac{3^n\left(n+1\right)}{3^{n+1}n}\\ =\underset{n\to \infty }{\lim }\frac{3^n\left(n+1\right)}{3^n\times 3n}\\ =\underset{n\to \infty }{\lim }\frac{n+1}{3n}\end{array}$

Further simplified as:

  $\begin{array}{c}r=\underset{n\to \infty }{\lim }\frac{1+\frac{1}{n}}{3}\\ =\frac{1}{3}\end{array}$

Since $r<1$ the series is convergent.