Chapter 8.3, Problem 4E

a.

To determine

To discuss: The behavior of the series ${\displaystyle \sum a_n}$ if $a_n>b_n$ for all $n$ .

Answer to Problem 4E

  ${\displaystyle \sum a_n}$ is divergent.

Explanation of Solution

Given Information: ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ are series with positive terms and ${\displaystyle \sum b_n}$ is divergent.

Since ${\displaystyle \sum b_n}$ is divergent, $\therefore \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{b}_k}=\infty$ .

  $\begin{array}{l}{a}_n>b_n\\ \\ \Rightarrow {\displaystyle \sum _{k=1}^n{a}_k>}{\displaystyle \sum _{k=1}^n{b}_k}\\ \\ \Rightarrow \underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n{a}_k=\infty }\end{array}$

Thus, ${\displaystyle \sum a_n}$ is divergent.

b.

To determine

To discuss: The behavior of the series ${\displaystyle \sum a_n}$ if $a_n<b_n$ for all $n$ .

Answer to Problem 4E

Convergence/divergence of ${\displaystyle \sum a_n}$ can’t be concluded.

Explanation of Solution

Given Information: ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ are series with positive terms and ${\displaystyle \sum b_n}$ is divergent.

Calculation:

Let $a_n=\frac{1}{n^2}$ and $b_n=\frac{1}{n}$ .

Then both $a_n$ and $b_n$ are positive for $n\ge 1$ .

Also, $\frac{1}{n^2}<\frac{1}{n}forn\ge 1$

  ${\displaystyle \sum _{n=1}^{\infty }{b}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ is divergent but ${\displaystyle \sum _{n=1}^{\infty }{a}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ is convergent.

Now, let $a_n=\frac{1}{2n}$ and $b_n=\frac{1}{n}$ .

Then both $a_n$ and $b_n$ are positive for $n\ge 1$ .

Also, $\frac{1}{2n}<\frac{1}{n}forn\ge 1$

Both ${\displaystyle \sum _{n=1}^{\infty }{b}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ and ${\displaystyle \sum _{n=1}^{\infty }{a}_n}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{2n}}$ are divergent.

Thus, convergence/divergence of ${\displaystyle \sum a_n}$ can’t be concluded.