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Found in: Page 443

### Essential Calculus: Early Transcendentals

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

# 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

## 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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