Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Find the value of \(c\) if \(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2} \)

\(c = \frac{{ - 1 + \sqrt 3 }}{2}\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Form reduction.

Given that this equation holds:\(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2} \).

This is of the form of a geometric series with ratio \({(1 + c)^{ - 1}}\)

\(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = {{(1 + c)}^{ - 2}}} + {(1 + c)^{ - 3}} + {(1 + c)^{ - 4}} + .... = {\left( {\frac{1}{{1 + c}}} \right)^2} + {\left( {\frac{1}{{1 + c}}} \right)^3} + {\left( {\frac{1}{{1 + c}}} \right)^4} + ...\)

\( = \sum\limits_{n = 1}^\infty {{{\left( {\frac{1}{{1 + c}}} \right)}^2}.{{\left( {\frac{1}{{1 + c}}} \right)}^{n - 1}}} \)

Comparison with geometric series

After comparison with \(\sum\limits_{n = 1}^\infty {a{r^{n - 1}}} \), we observe that \(a = {\left( {\frac{1}{{1 + c}}} \right)^2},r = \frac{1}{{1 + c}}\).

Using geometric formula further, \(\begin{aligned}\sum\limits_{n = 1}^\infty {a{r^{n - 1}}} & = \frac{a}{{1 - r}} = \sum\limits_{n = 1}^\infty {{{\left( {\frac{1}{{1 + c}}} \right)}^2}.} {\left( {\frac{1}{{1 + c}}} \right)^{n - 1}} = \frac{{{{\left( {\frac{1}{{1 + c}}} \right)}^2}}}{{1 - \frac{1}{{1 + c}}}} = \frac{1}{{{{(1 + c)}^2}}} \times \frac{{(1 + c)}}{c}\\ = \frac{1}{{c + {c^2}}}\end{aligned}\)

Now we combine this obtained result with \(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2} \)(given).

\(2 = \frac{1}{{c + {c^2}}} \Rightarrow 2{c^2} + 2c - 1 = 0\)

Solving quadratic formula

Solving using quadratic formula, we obtain:

\(c = \frac{{ - 2 \pm \sqrt {{{(2)}^2} - 4(2)( - 1)} }}{{2(2)}} = \frac{{ - 2 \pm \sqrt {4 + 8} }}{4} = \frac{{ - 2 \pm 2\sqrt 3 }}{4} = \frac{{ - 1 \pm \sqrt 3 }}{2}\)

Now for the given series to be convergent geometric series \(|r| < 1 \Rightarrow |\frac{1}{{1 + c}}| < 1 \Rightarrow |1 + c| > 1\).

We also see that \(\frac{{ - 1 + \sqrt 3 }}{2} = 0.366{\rm{ and }}\frac{{ - 1 - \sqrt 3 }}{2} = - 1.366\).

Hence we conclude that \(c = \frac{{ - 1 + \sqrt 3 }}{2}\)since \(|r| < 1 \Rightarrow |\frac{1}{{1 + c}}| < 1 \Rightarrow |1 + c| > 1\) holds true for this value of \(c\).

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Essential Calculus: Early Transcendentals

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Short Answer

Find the value of \(c\) if \(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2} \)

Step by Step Solution

Form reduction.

Comparison with geometric series

Solving quadratic formula

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Find the value of \(c\) if \(\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2} \)

Step by Step Solution

Form reduction.

Comparison with geometric series

Solving quadratic formula

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