Chapter 8.3, Problem 24E

To determine

To find: $\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\mathrm{...}$is convergent or not .

Answer to Problem 24E

  $\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\mathrm{...}$ is divergent .

Explanation of Solution

Given information:

  $\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\mathrm{...}$

Calculation:

  $\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\frac{1}{17}+\mathrm{...}$

  $={\displaystyle \sum _{n=1}^{\infty }\frac{1}{3n+2}}$

Let , ${\displaystyle \sum a_n={\displaystyle \sum \frac{1}{n}}}$and ${\displaystyle \sum b_n={\displaystyle \sum \frac{1}{3n+2}}}$

Then by limit form :

  $\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{1}{n}.3n+2\\ =\underset{n\to \infty }{\lim }(3+\frac{2}{n})\\ \end{array}$

  $=3$ (positive real number)

Hence , ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ coverge or diverge together .

Since , ${\displaystyle \sum \frac{1}{n}}$ is divergent tjen by comparison test ${\displaystyle \sum \frac{1}{3n+2}}$ is divergent .

Hence , $\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\mathrm{...}$ is divergent .