Chapter 10.5, Problem 65E

  $a$

To determine

To find: which one of the series fails to satisfy the condition of alternating series test.

Answer to Problem 65E

It fails to satisfy $u_n\ge u_{n+1}$ for all $n\ge N$

Explanation of Solution

Given information:

Given series are : $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\dots \dots +\frac{1}{3^n}-\frac{1}{2^n}+\dots \dots$

Calculation:

The series is

  $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\dots \dots +\frac{1}{3^n}-\frac{1}{2^n}+\dots \dots$

This can be represented as

  $\left({\displaystyle \sum }_{n=1}^{\infty }\frac{1}{3^n}\right)-\left({\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2^n}\right)$

It fails to satisfy $u_n\ge u_{n+1}$ for all $n\ge N$

  $b$

To determine

To find : sum of the series in part.

Answer to Problem 65E

The sum of the series in part (a) $=-\frac{1}{2}$

Explanation of Solution

Given information:

Given series are : $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\dots \dots +\frac{1}{3^n}-\frac{1}{2^n}+\dots \dots$

Calculation:

The series is combination of two geometric progressions and we know that the sum of a G.P. is given by $\frac{a}{1-r}$

so the sum is

  $\begin{array}{cc}\left({\displaystyle \sum }_{n=1}^{\infty }\frac{1}{3^n}\right)-\left({\displaystyle \sum }_{n=1}^{\infty }\frac{1}{2^n}\right)& =\frac{1/3}{1-1/3}-\frac{1/2}{1-1/2}\\ & =\frac{1}{2}-1\\ & =-\frac{1}{2}\end{array}$