Chapter 9.5, Problem 65E

a.

To determine

To find: To find whether it satisfy the given condition.

Answer to Problem 65E

It fails to satisfy.

Explanation of Solution

Given:

The given series $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\mathrm{.......}+\frac{1}{3^n}-\frac{1}{2^n}$ .

Formula used:

Sum is given for an infinite geometric progression is given by $S=\frac{a}{1-r}$ where a is first and r is common ratio, also $-1<r<1$ .

Calculation:

Given series is $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\mathrm{.......}+\frac{1}{3^n}-\frac{1}{2^n}$ .

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: To find the sum of the series.

Answer to Problem 65E

The sum of series is $-\frac{1}{2}$ .

Explanation of Solution

Given:

The given series $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\mathrm{.......}+\frac{1}{3^n}-\frac{1}{2^n}$ .

Formula used:

Sum is given for an infinite geometric progression is given by $S=\frac{a}{1-r}$ where a is first and r is common ratio, also $-1<r<1$ .

Calculation:

The given series $\frac{1}{3}-\frac{1}{2}+\frac{1}{9}-\frac{1}{4}+\frac{1}{27}-\frac{1}{8}+\mathrm{.......}+\frac{1}{3^n}-\frac{1}{2^n}$ .

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