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Essential Calculus: Early Transcendentals
Found in: Page 443
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

Determine whether the geometric series is convergent or divergent..If it is convergent,find its sum.

\(\sum\limits_{n = 1}^\infty {\frac{{{{( - 3)}^{n - 1}}}}{{{4^n}}}} \)

The given series is Convergent and Sum=\(\frac{1}{7}\)..\(\)

See the step by step solution

Step by Step Solution

Finding the common ratio

Given Series \(\sum\limits_{n = 1}^\infty {\frac{{{{( - 3)}^{n - 1}}}}{{{4^n}}}} = \frac{1}{4} - \frac{3}{{16}} + \frac{9}{{64}} - \frac{{27}}{{256}} + ......\)

Common Ratio of this series=\(\frac{{{a_{n + 1}}}}{{{a_n}}}\)=\(\frac{{ - 3}}{{16}}*\frac{4}{1} = - \frac{3}{4}\)

Checking convergent

Since,\(|r| = \frac{3}{4} < 1\)

Therefore,the series is convergent..

Finding Sum..

Formula for infinite sum of a gp is =\(\frac{a}{{1 - r}}\)

Where a is first term and r is common ratio.

\({S_\infty } = \frac{{\frac{1}{4}}}{{1 + \frac{3}{4}}} = \frac{1}{7}\)

Hence,The Series is convergent and Sum=\(\frac{8}{3}\)

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