Problem 64
Question
Find each sum that converges. $$\sum_{i=1}^{\infty}\left(-\frac{1}{3}\right)\left(\frac{3}{4}\right)^{i-1}$$
Step-by-Step Solution
Verified Answer
The series converges to -4/3.
1Step 1: Identify the Series Type
This is an infinite geometric series. A geometric series has the general form \( \sum_{i=1}^{\infty} ar^{i-1} \), where \( a \) is the first term and \( r \) is the common ratio of the series.
2Step 2: Determine the First Term and Common Ratio
From the given series \( \sum_{i=1}^{\infty}\left(-\frac{1}{3}\right)\left(\frac{3}{4}\right)^{i-1} \): - The first term \( a \) is \(-\frac{1}{3} \).- The common ratio \( r \) is \( \frac{3}{4} \).
3Step 3: Check Convergence
A geometric series converges if and only if the absolute value of the common ratio is less than 1, i.e., \(|r| < 1\). In this series, \( r = \frac{3}{4} \), and since \( \left| \frac{3}{4} \right| = 0.75 < 1 \), the series converges.
4Step 4: Use the Formula for the Sum of a Convergent Geometric Series
The sum \( S \) of an infinite convergent geometric series is calculated using the formula: \[ S = \frac{a}{1 - r} \]
5Step 5: Calculate the Sum of the Series
Substitute \( a = -\frac{1}{3} \) and \( r = \frac{3}{4} \) into the formula: \[ S = \frac{-\frac{1}{3}}{1 - \frac{3}{4}} = \frac{-\frac{1}{3}}{\frac{1}{4}} = -\frac{1}{3} \times 4 = -\frac{4}{3} \]The sum of the series is \(-\frac{4}{3}\).
Key Concepts
Infinite SeriesConvergent SeriesSum of SeriesCommon Ratio
Infinite Series
An infinite series is a sum of infinitely many terms, typically written in the form of a sequence. Each term in the series contributes to the total sum, and when visualized, this process continues endlessly. In practical terms, we're often interested in whether this sum reaches a finite value. Infinite series appear in various mathematical scenarios, from simple arithmetic to complex calculus functions.
- **Representation**: The infinite series is represented as \( \sum_{i=1}^{\infty} a_i \), where \( a_i \) represents each term in the series.
- **Importance**: Infinite series are foundational in mathematical concepts like power series, Taylor series, and Fourier series, which have widespread applications in both pure and applied mathematics.
Convergent Series
A convergent series is a type of infinite series where the sum of its terms approaches a finite limit. This means that as you add more and more terms, the total sum gets closer to a specific finite number. This is an important concept because not all infinite series converge.
- **Criteria for Convergence**: For a geometric series, convergence is determined by the common ratio \( |r| < 1 \). This condition ensures that the terms in the series diminish in size, allowing the sum to stabilize towards a fixed value.
- **Why It Matters**: Convergence indicates that the infinite series leads to meaningful conclusions in applications like physics and engineering, where calculations rely on obtaining concrete results from series approximations.
Sum of Series
Once we've identified that a series is convergent, the next step is finding its sum. For geometric series, this involves a straightforward formula that helps us calculate the total quickly and efficiently.
- **Geometric Series Sum Formula**: For an infinite geometric series with first term \( a \) and common ratio \( r \), the sum \( S \) is given by: \[ S = \frac{a}{1 - r} \]
- **Example Calculation**: Using this formula with our series, where \( a = -\frac{1}{3} \) and \( r = \frac{3}{4} \), we substitute and find \( S = -\frac{4}{3} \).
Common Ratio
The common ratio in a geometric series is the factor by which each subsequent term is multiplied to get the next term. It's an essential part of defining the series' nature and behavior.
- **Identification**: In the series \( \sum_{i=1}^{\infty}\left(-\frac{1}{3}\right)\left(\frac{3}{4}\right)^{i-1} \), the common ratio \( r \) is \( \frac{3}{4} \).
- **Role in Convergence**: The convergence of a series depends greatly on the absolute value of the common ratio being less than one (\( |r| < 1 \)). This determines whether extra terms keep the sum from becoming infinite.
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