Determine whether the series converges or diverges. If it converges, find the sum.3" – 1(Hint: The summation of a difference is the difference of the5"n=1summations.)

Answered: Determine whether the series converges…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Determine whether the series converges or diverges. If it converges, find the sum.
3" – 1
(Hint: The summation of a difference is the difference of the
5"
n=1
summations.)

Expert Solution

Step 1

Given: $S = \sum_{n = 1}^{\infty} \frac{3^{n} - 1}{5^{n}}$

for finding given summation, we expand given summation by putting value of $n = 1, 2, 3, . . . . .$

so,

$\begin{array}{rcl} S & = & \sum_{n = 1}^{\infty} \frac{3^{n} - 1}{5^{n}} \\ = & \sum_{n = 1}^{\infty} {(\frac{3}{5})}^{n} - \sum_{n = 1}^{\infty} {(\frac{1}{5})}^{n} \\ = & (\frac{3}{5} + {(\frac{3}{5})}^{2} + {(\frac{3}{5})}^{3} + . . . . . . . . . .) - (\frac{1}{5} + {(\frac{1}{5})}^{2} + {(\frac{1}{5})}^{3} + . . . . . . . . . .) . . (1) \end{array}$

now, we can see that given sequence is infinite geometric sequence whose first terms are $\frac{3}{5}, \frac{1}{5}$ and common ratio are $\frac{3}{5}, \frac{1}{5}$

Step 2

we know that sum of infinite geometric sequence whose first term is a and common ratio is r given by

$S = \frac{a}{1 - r} . . (2)$

similarly, by using equation $(2)$ sum of equation $(1)$ will be

$\begin{array}{rcl} S & = & \frac{(\frac{3}{5})}{1 - \frac{3}{5}} - \frac{(\frac{1}{5})}{1 - \frac{1}{5}} \\ = & \frac{(\frac{3}{5})}{(\frac{2}{5})} - \frac{(\frac{1}{5})}{(\frac{4}{5})} \\ = & (\frac{3}{5}) (\frac{5}{2}) - (\frac{1}{5}) (\frac{5}{4}) \\ = & \frac{3}{2} - \frac{1}{4} \\ = & \frac{3 (2) - 1}{4} \\ = & \frac{6 - 1}{4} \\ = & \frac{5}{4} \end{array}$

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