Problem 32 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$ \displaystyle \sum_{n = 1}^{\infty} [(-0.2)^n + (0.06)^{n - 1}] $$

Step-by-Step Solution

Answer

The series converges, and its sum is \( \frac{253}{282} \).

Step 1: Break Down the Series

The given series can be expressed as two separate series:\[ \sum_{n=1}^{\infty} (-0.2)^n + \sum_{n=1}^{\infty} (0.06)^{n-1} \]We will analyze each series individually to determine convergence.

Step 2: Analyze the First Series

Consider the series \( \sum_{n=1}^{\infty} (-0.2)^n \). This is a geometric series with the first term \( a = -0.2 \) and common ratio \( r = -0.2 \). A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if \( |r| < 1 \). In this case, \(|r| = 0.2 < 1\), so the series converges.

Step 3: Find the Sum of the First Series

For a geometric series \( \sum_{n=1}^{\infty} ar^n \), the sum is given by \( S = \frac{a}{1-r} \) when \( |r| < 1 \). Here, \( a = -0.2 \) and \( r = -0.2 \). Therefore:\[ S = \frac{-0.2}{1 - (-0.2)} = \frac{-0.2}{1.2} = -\frac{1}{6} \]

Step 4: Analyze the Second Series

Consider the series \( \sum_{n=1}^{\infty} (0.06)^{n-1} \). Rewrite it as \( \sum_{n=0}^{\infty} (0.06)^n \), which is a geometric series with \( a = 1 \) and \( r = 0.06 \).The series converges because \( |r| = 0.06 < 1 \).

Step 5: Find the Sum of the Second Series

Since it is a geometric series with first term \( a = 1 \) and common ratio \( r = 0.06 \), the sum is given by:\[ S = \frac{a}{1-r} = \frac{1}{1-0.06} = \frac{1}{0.94} = \frac{50}{47} \]

Step 6: Combine the Results

To find the total sum of the original series, add the sums of the two component series:\[ S_{total} = -\frac{1}{6} + \frac{50}{47} \]To add these, find a common denominator. The least common denominator is 282. Therefore:\[ S_{total} = \frac{-47}{282} + \frac{300}{282} = \frac{253}{282} \]

Step 7: Conclusion

The original series converges, and its sum is calculated as \( \frac{253}{282} \).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. The series can be expressed in the form:

• \(S = a + ar + ar^2 + ar^3 + \ldots = \sum_{n=0}^{\infty} ar^n\)

where \(a\) is the first term and \(r\) is the common ratio.
This series can be either finite or infinite. In infinite cases, whether the series converges or not depends on the common ratio's absolute value.
In our given exercise, the first component series \((-0.2)^n\) and the second component series \((0.06)^{n-1}\) are both geometric.

Common Ratio

The common ratio in a geometric series is the constant factor between consecutive terms. It can determine the behavior of the series, particularly its convergence or divergence.
Mathematically, the common ratio \(r\) is defined as:

• \(r = \frac{a_{n+1}}{a_n}\)

For example, in the series \((-0.2)^n\), the common ratio is \(-0.2\).
In the series \((0.06)^{n-1}\), the common ratio is \(0.06\).
These series are special because their common ratios are both less than 1 in absolute value, which is a key factor in their convergence.

Series Convergence

A series is said to converge if the sum of its terms approaches a finite number as more terms are added. For geometric series, convergence depends critically on the value of the common ratio \(r\).
The series \(\sum_{n=0}^{\infty} ar^n\) converges if and only if \(|r| < 1\).
This means the series terms become smaller and smaller, eventually negligible, allowing a finite sum.
In our exercise, both series \(\sum_{n=1}^{\infty} (-0.2)^n\) and \(\sum_{n=0}^{\infty} (0.06)^n\) converge because their common ratios \(0.2\) and \(0.06\), respectively, are less than 1 in absolute value.

Sum of Series

For a convergent geometric series where \(|r| < 1\), the sum can be calculated using a specific formula:

• \[ S = \frac{a}{1-r} \]

where \(a\) is the first term and \(r\) is the common ratio. This formula arises because the terms get smaller geometrically, and their sum approaches a fixed point.
In the provided exercise, the sum of the first series \((-0.2)^n\) is:

• \(\frac{-0.2}{1 - (-0.2)} = -\frac{1}{6}\).

And, the sum of the second series \((0.06)^{n-1}\) is:

• \(\frac{1}{1-0.06} = \frac{50}{47}\).

When you combine these sums to find the total sum of the original series, you use:

• \(S_{total} = -\frac{1}{6} + \frac{50}{47} = \frac{253}{282}\)