Problem 15
Question
In Problems 15-20, determine whether the given geometric series is convergent or divergent. If convergent, find its sum. $$ \sum_{k=0}^{\infty}(1-i)^{k} $$
Step-by-Step Solution
Verified Answer
The series diverges.
1Step 1: Identify the form of the series
The series is given as \( \sum_{k=0}^{\infty}(1-i)^{k} \). This is a geometric series of the form \( a + ar + ar^2 + \ldots \), where \( a \) is the first term and \( r \) is the common ratio. In this series, \( a = 1 \) and \( r = 1-i \).
2Step 2: Check the condition for convergence
A geometric series \( \sum_{k=0}^{\infty} ar^k \) converges if and only if the absolute value of the common ratio \( |r| < 1 \). Our common ratio is \( r = 1-i \).
3Step 3: Calculate the magnitude of the common ratio
To find \(|1-i|\), calculate the magnitude:\[|1-i| = \sqrt{(1)^2 + (-1)^2} = \sqrt{2}\]Since \(\sqrt{2} > 1\), the magnitude of \(1-i\) is greater than 1.
4Step 4: Determine the convergence or divergence
Since \(|1-i| = \sqrt{2} > 1\), the series \( \sum_{k=0}^{\infty}(1-i)^{k} \) does not satisfy the convergence condition for a geometric series. Thus, the series diverges.
Key Concepts
Convergence of SeriesComplex NumbersDivergent Series
Convergence of Series
When exploring the convergence of a series, it's important to understand when a series will settle at a specific value or continue endlessly. In mathematics, a series converges if the sum of its infinite terms approaches a fixed value as more terms are added. For a geometric series, which is a series of numbers where each term is a fixed multiple of the previous one, convergence depends on the common ratio, denoted as \( r \).
The essential rule for convergence is straightforward: a geometric series \( \sum_{k=0}^{\infty} ar^k \) converges if the absolute value of the common ratio \( |r| \) is less than 1. This means that as k increases, the terms \( ar^k \) become smaller and smaller in absolute value, allowing the overall sum to approach a finite limit.
The essential rule for convergence is straightforward: a geometric series \( \sum_{k=0}^{\infty} ar^k \) converges if the absolute value of the common ratio \( |r| \) is less than 1. This means that as k increases, the terms \( ar^k \) become smaller and smaller in absolute value, allowing the overall sum to approach a finite limit.
- If \( |r| < 1 \), the series converges, and you can find the sum using the formula \( \frac{a}{1-r} \), where \( a \) is the first term of the series.
- If \( |r| \geq 1 \), the series does not converge, meaning it is divergent.
Complex Numbers
Complex numbers are numbers that have both a real part and an imaginary part. The standard form is \( a + bi \), where \( a \) is the real component and \( bi \) is the imaginary component. Imaginary numbers use \( i \), which is the square root of -1.
When working with series, complex numbers introduce unique conditions for analyzing convergence or divergence. In geometric series, the common ratio \( r \) can be a complex number. The convergence criteria still apply, but you need to consider the magnitude (or absolute value) of the complex number.
The magnitude of a complex number \( a + bi \) is calculated using the formula \( |a + bi| = \sqrt{a^2 + b^2} \). This helps in determining if a series will converge when such numbers are involved. For instance, in the series \( \sum_{k=0}^{\infty} (1-i)^k \), the common ratio \( r = 1-i \) has a magnitude that influences the convergence analysis.
When working with series, complex numbers introduce unique conditions for analyzing convergence or divergence. In geometric series, the common ratio \( r \) can be a complex number. The convergence criteria still apply, but you need to consider the magnitude (or absolute value) of the complex number.
The magnitude of a complex number \( a + bi \) is calculated using the formula \( |a + bi| = \sqrt{a^2 + b^2} \). This helps in determining if a series will converge when such numbers are involved. For instance, in the series \( \sum_{k=0}^{\infty} (1-i)^k \), the common ratio \( r = 1-i \) has a magnitude that influences the convergence analysis.
Divergent Series
A series is said to be divergent if it does not converge to a limit as more terms are added. In mathematical terms, a divergent series will have terms that either grow without bound or do not settle to a definite number.
For geometric series, the condition for divergence is tied directly to the value of the common ratio \( r \). If the magnitude \( |r| \geq 1 \), the series is divergent. This means the terms are not decreasing in a manner that they sum to a finite value, leading the series to drift towards infinity or oscillate.
For geometric series, the condition for divergence is tied directly to the value of the common ratio \( r \). If the magnitude \( |r| \geq 1 \), the series is divergent. This means the terms are not decreasing in a manner that they sum to a finite value, leading the series to drift towards infinity or oscillate.
- This characteristic affects both real and complex numbers in the series.
- In our example, the series \( \sum_{k=0}^{\infty}(1-i)^k \) is divergent because the common ratio \( |1-i| = \sqrt{2} > 1 \).
- Since it does not meet the convergence condition, the sum cannot be determined.
Other exercises in this chapter
Problem 15
Expand \(f(z)=\frac{1}{(z-1)(z-2)}\) in a Laurent series valid for the given annular domain. $$ 0
View solution Problem 15
In Problems 15-22, expand the given function in a Taylor series centered at the indicated point \(z_{0}\). Give the radius of convergence \(R\) of each series.
View solution Problem 16
Evaluate the Cauchy principal value of the given improper integral. $$ \int_{-\infty}^{\infty} \frac{1}{x^{2}-6 x+25} d x $$
View solution Problem 16
Determine the order of the poles for the given function. $$ f(z)=5-\frac{6}{z^{2}} $$
View solution