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Q20E

Expert-verifiedFound in: Page 443

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.

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.

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

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**

\({\sum\limits_{n = 1}^\infty {(0.8)} ^{n - 1}}\)** **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** ****\({\sum\limits_{n = 1}^\infty {(0.8)} ^{n - 1}} = 5\)**** **

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.

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

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