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Magazine Chapter 11: Problem 32
Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$ \sum_{n=1}^{\infty}\left[(-0.2)^{n}+(0.6)^{n-1}\right] $$
Short Answer
Expert verified
The series is convergent with a sum of \( \frac{7}{3} \).
Step by step solution
01
Identify the Series Components
The series given is \( \sum_{n=1}^{\infty} \left[ (-0.2)^{n} + (0.6)^{n-1} \right] \). We can split this into two separate series: \( \sum_{n=1}^{\infty} (-0.2)^n \) and \( \sum_{n=1}^{\infty} (0.6)^{n-1} \). Each component is a geometric series.
02
Determine Convergence of First Series
The series \( \sum_{n=1}^{\infty} (-0.2)^n \) is a geometric series with a common ratio of \(-0.2\). Since the absolute value of the common ratio is less than 1, \(|-0.2| = 0.2 < 1\), the series converges.
03
Sum of the First Geometric Series
The sum of a convergent geometric series \( \sum_{n=1}^{\infty} ar^n \) is \( \frac{ar}{1-r} \), where \( a \) is the first term. For \( \sum_{n=1}^{\infty} (-0.2)^n \), \( a = (-0.2)^1 = -0.2 \) and \( r = -0.2 \). Thus, the sum is \[ \frac{-0.2}{1+0.2} = \frac{-0.2}{1.2} = -\frac{1}{6}. \]
04
Determine Convergence of Second Series
The series \( \sum_{n=1}^{\infty} (0.6)^{n-1} \) can be rewritten starting at \( n = 1 \) by changing the index, \( \sum_{n=0}^{\infty} (0.6)^n \), which is a geometric series with a common ratio of \(0.6\). Since \(|0.6| < 1\), this series also converges.
05
Sum of the Second Geometric Series
For the series \( \sum_{n=0}^{\infty} (0.6)^n \), the first term is \( a = 1 \) and the common ratio is \( r = 0.6 \). The sum is \[ \frac{1}{1-0.6} = \frac{1}{0.4} = 2.5. \]
06
Find the Sum of the Original Series
Since both series converge, their combined sum \( \sum_{n=1}^{\infty} \left[ (-0.2)^n + (0.6)^{n-1} \right] \) is the sum of the individual series: \(-\frac{1}{6} + 2.5\). Calculating this gives \[ 2.5 - \frac{1}{6} = \frac{15}{6} - \frac{1}{6} = \frac{14}{6} = \frac{7}{3}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Convergence
When dealing with series, especially infinite series, determining whether a series converges or diverges is fundamental. Convergence means the series approaches a specific value as more terms are added, while divergence means it does not settle at any particular number.
For a geometric series, this largely depends on the common ratio, denoted by \( r \). A geometric series of the form \( \sum_{n=0}^{\infty} ar^n \) will converge if the absolute value of \( r \) is less than 1, that is \( |r| < 1 \). If \( |r| \geq 1 \), the series diverges.
In our original exercise, we had two geometric series: \( \sum_{n=1}^{\infty} (-0.2)^n \) and \( \sum_{n=1}^{\infty} (0.6)^{n-1} \). Both of these series converge because the absolute values of their common ratios, \(-0.2\) and \(0.6\), are less than 1.
For a geometric series, this largely depends on the common ratio, denoted by \( r \). A geometric series of the form \( \sum_{n=0}^{\infty} ar^n \) will converge if the absolute value of \( r \) is less than 1, that is \( |r| < 1 \). If \( |r| \geq 1 \), the series diverges.
In our original exercise, we had two geometric series: \( \sum_{n=1}^{\infty} (-0.2)^n \) and \( \sum_{n=1}^{\infty} (0.6)^{n-1} \). Both of these series converge because the absolute values of their common ratios, \(-0.2\) and \(0.6\), are less than 1.
Sum of Series
The sum of a convergent geometric series can be calculated using a straightforward formula, which is particularly useful. This formula is \( \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio of the series.
For the first series in the exercise, \( \sum_{n=1}^{\infty} (-0.2)^n \), the first term \( a \) is \(-0.2\), so the sum is calculated as:
For the first series in the exercise, \( \sum_{n=1}^{\infty} (-0.2)^n \), the first term \( a \) is \(-0.2\), so the sum is calculated as:
- \( \frac{-0.2}{1+0.2} = -\frac{1}{6} \).
- \( \frac{1}{1-0.6} = 2.5 \).
- \( -\frac{1}{6} + 2.5 = \frac{7}{3} \).
Common Ratio
The common ratio in a geometric series is a crucial aspect that helps determine both its convergence and method of finding the sum. The common ratio, \( r \), is obtained by dividing any term in the series by the previous term.
In the exercise at hand, for the series \( \sum_{n=1}^{\infty} (-0.2)^n \), the common ratio is \(-0.2\). This negative value is still valid; what matters for convergence is the absolute value \( |r| \).
In the exercise at hand, for the series \( \sum_{n=1}^{\infty} (-0.2)^n \), the common ratio is \(-0.2\). This negative value is still valid; what matters for convergence is the absolute value \( |r| \).
- If \( |r| < 1 \), the series converges.
- If \( |r| \geq 1 \), the series diverges.
