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Calculus / Calculus: Early Transcendentals 1 / Chapter 8.3 / Problem 81AE

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Series in an equation For what values of x does the

Chapter 11, Problem 81AE

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QUESTION:

Series in an equation For what values of x does the geometric series

$f(x)=\sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}$

converge? Solve f(x) = 3.

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QUESTION:

Series in an equation For what values of x does the geometric series

$f(x)=\sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}$

converge? Solve f(x) = 3.

ANSWER:

Problem 81AE

Series in an equation For what values of x does the geometric series

$f(x)=\sum\limits_{k\ =\ 0}^{\infty{}}(\frac{1}{1+x}{)}^{k}$ converge? Solve f(x) = 3.

Solution $\$

Step 1

In this problem we have to evaluate f(3) where $f(x)=\sum\limits_{k\ =\ 0}^{\infty{}}(\frac{1}{1+x}{)}^{k}$

We know that “If $|r|\ <1$ then the sum of the infinite geometric series is $\sum\limits_{k\ =\ 0}^{\infty{}}a{r}^{k}=\frac{a}{1\ -\ r}.$ If $|r|\geq{}1$ then the series diverges.”

We have $f(x)=\sum\limits_{k\ =\ 0}^{\infty{}}(\frac{1}{1+x}{)}^{k}$

Here $a=1,r=\frac{1}{1+x}$

Hence $f(x)=\sum\limits_{k\ =\ 0}^{\infty{}}(\frac{1}{1+x}{)}^{k}=\frac{1}{1\ -\ \frac{1}{1+x}}\ =\frac{x+1}{x}$ if $\frac{1}{\left|{x+1}\right|}\ <1$

Therefore by geometric series test the series is converges for $\frac{1}{\left|{x+1}\right|}\ <1$

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Series in an equation For what values of x does the

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Solution Found!

Series in an equation For what values of x does the

Questions & Answers

Study Tools You Might Need

Chapter: 7 Problem: 74P

Chapter: 16 Problem: 96P

Chapter: 24 Problem: 2DQ

Chapter: 8 Problem: 51P

Chapter: Problem: 12.53

Chapter: 5 Problem: 32

Chapter: 18 Problem: 4

Chapter: 9 Problem: 7

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