Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:**

(1) Find the summation of the series

\[
\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots
\]

**Explanation:**

This is an infinite geometric series where the first term \(a\) is \(\frac{1}{3}\) and the common ratio \(r\) is also \(\frac{1}{3}\).

The sum \(S\) of an infinite geometric series can be found using the formula:

\[
S = \frac{a}{1 - r}
\]

where \(|r| < 1\).

Applying the values from the series:

- \(a = \frac{1}{3}\)
- \(r = \frac{1}{3}\)

Thus,

\[
S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
\]

Therefore, the summation of the series is \(\frac{1}{2}\).
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Transcribed Image Text:**Problem Statement:** (1) Find the summation of the series \[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \] **Explanation:** This is an infinite geometric series where the first term \(a\) is \(\frac{1}{3}\) and the common ratio \(r\) is also \(\frac{1}{3}\). The sum \(S\) of an infinite geometric series can be found using the formula: \[ S = \frac{a}{1 - r} \] where \(|r| < 1\). Applying the values from the series: - \(a = \frac{1}{3}\) - \(r = \frac{1}{3}\) Thus, \[ S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \] Therefore, the summation of the series is \(\frac{1}{2}\).
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