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Found in: Page 443

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# :$$\sum\limits_{n = 1}^\infty {\left( {{{(0.8)}^{n - 1}} - {{(0.3)}^n}} \right)}$$ Find Whether It Is Convergent Or Divergent. If It Is Convergent Find Its Sum.

Given $$\sum\limits_{n = 1}^\infty {\left( {{{(0.8)}^{n - 1}} - {{(0.3)}^n}} \right)}$$

The Summation is Difference of Two Geometric Series.

See the step by step solution

## Rearrange $$\sum\limits_{n = 1}^\infty {\left( {{{(0.8)}^{n - 1}} - {{(0.3)}^n}} \right)}$$

By series Difference/Sum Rule, if $$\sum {{a_n}}$$ and $$\sum {{b_n}}$$ both are Convergent Series then$$\sum {{a_n}} \pm {b_n} = \sum {{a_n} \pm \sum {{b_n}} }$$.

If any one of them are divergent $$\sum {{a_n}} \pm {b_n} = \sum {{a_n}} \pm \sum {{b_n}}$$ may or may not be possible.

## Assumption

Let $$\sum\limits_{n = 1}^\infty {\left( {{{(0.8)}^{n - 1}} + {{(0.3)}^n}} \right)} = \sum\limits_{n = 1}^\infty {{{(0.8)}^{n - 1}}} - \sum\limits_{n = 1}^\infty {{{(0.3)}^n}}$$

## Finding $$\sum\limits_{n = 1}^\infty {{{(0.8)}^{n - 1}}}$$  Is Convergent Or Divergent

Consider $$\sum\limits_{n = 1}^\infty {{{(0.8)}^{n - 1}}} = {(0.8)^{1 - 1}} + {(0.8)^{2 - 1}} + {(0.8)^{3 - 1}} + ................$$

\begin{aligned} &= {(0.8)^0} + {(0.8)^1} + {(0.8)^2} + ...........\\ &= {(\frac{8}{{10}})^0} + {(\frac{8}{{10}})^1} + {(\frac{8}{{10}})^2} + .............\end{aligned}

$$\sum\limits_{n = 1}^\infty {{{(0.8)}^{n - 1}}} = 1 + (\frac{8}{{10}}) + {(\frac{8}{{10}})^2} + .........$$

It is a Geometric Series with $$r = \frac{8}{{10}}$$ and $$a = 1$$ ; where “a” is the First Term in Geometric Series and “r” is the Common Ratio.

$$\left| {\frac{8}{{10}}} \right| < 1$$ i.e $$- 1 < \frac{8}{{10}} < 1$$ here $$- 1 < r < 1$$

Therefore it is a Convergent Series

## Finding the $$\sum\limits_{n = 1}^\infty {{{(0.8)}^{n - 1}}}$$ Value

$${\sum\limits_{n = 1}^\infty {(0.8)} ^{n - 1}}$$ is a Geometric Series and Convergent ,Hence The Series Converges to $$\frac{a}{{1 - r}} = \frac{1}{{1 - \frac{8}{{10}}}} \Rightarrow \frac{a}{{1 - r}} = \frac{{10}}{2}$$

$$\Rightarrow \frac{a}{{1 - r}} = 5$$

Therefore $${\sum\limits_{n = 1}^\infty {(0.8)} ^{n - 1}}$$ is a Convergent Series and Converges to 5 i.e $${\sum\limits_{n = 1}^\infty {(0.8)} ^{n - 1}} = 5$$

## Finding Whether $$\sum\limits_{n = 1}^\infty {{{(0.3)}^n}}$$is Convergent Or Divergent

Consider $$\sum\limits_{n = 1}^\infty {{{(0.3)}^n}} = {(0.3)^1} + {(0.3)^2} + {(0.3)^3} + ...........$$

$$= {(\frac{3}{{10}})^1} + {(\frac{3}{{10}})^2} + {(\frac{3}{{10}})^3} + .......$$

The Infinite Series $${\sum\limits_{n = 1}^\infty {(0.3)} ^n}$$ is a Geometric Series and $$a = (\frac{3}{{10}})$$, $$r = (\frac{3}{{10}})$$ . Here $$\left| {\frac{3}{{10}}} \right| < 1$$ i.e $$- 1 < \frac{3}{{10}} < 1$$ which means “r” lies between -1 and 1.Hence ,$$\sum\limits_{n = 1}^\infty {{{(0.3)}^n}}$$ is a Convergent Series.

## Finding $$\sum\limits_{n = 1}^\infty {{{(0.3)}^n}}$$ Value

The $${\sum\limits_{n = 1}^\infty {(0.3)} ^n}$$ is a Geometric Series and Converging Series .The Series Converges to $$\frac{a}{{1 - r}}$$

$$\Rightarrow \frac{a}{{1 - r}} = \frac{{(\frac{3}{{10}})}}{{\left( {1 - (\frac{3}{{10}})} \right)}}$$

$$= \frac{{ (\frac{3}{{10}})}}{{(\frac{7}{{10}})}}$$

$$(\frac{a}{{1 - r}}) = (\frac{3}{7})$$

Therefore the $${\sum\limits_{n = 1}^\infty {(0.3)} ^n} = \frac{3}{7}$$

Hence,$$\sum\limits_{n = 1}^\infty {\left( {{{\left( {0.8} \right)}^{n - 1}} - {{\left( {0.3} \right)}^n}} \right)} = \sum\limits_{n = 1}^\infty {{{\left( {o.8} \right)}^{n - 1}}} - \sum\limits_{n = 1}^\infty {{{\left( {0.3} \right)}^n}}$$

\begin{aligned} &= 5 - \frac{3}{7}\\ &= \frac{{35 - 3}}{7}\\ &= \frac{{32}}{7}\\ &= 4.\overline {571428} \end{aligned}

Hence, the $$\sum\limits_{n = 1}^\infty {\left( {{{\left( {0.8} \right)}^{n - 1}} - {{\left( {0.3} \right)}^n}} \right)} = 4.\overline {571428}$$ and it is a Converging Series