Problem 16 - A First Course in Complex Analysis with Applications

Question

Determine whether the given geometric series is convergent or divergent. If convergent, find its sum. $$ \sum_{k=1}^{\infty} 4 i\left(\frac{1}{3}\right)^{k-1} $$

Step-by-Step Solution

Answer

The series is convergent with a sum of \( 6i \).

Step 1: Identify the First Term

In the given geometric series, identify the first term. The series is written as \( \sum_{k=1}^{\infty} a r^{k-1} \). Here, \( a = 4i \) is the first term.

Step 2: Identify the Common Ratio

The common ratio \( r \) in the series is the factor that each term is multiplied by to get the next term. From the series \( \sum_{k=1}^{\infty} 4i\left(\frac{1}{3}\right)^{k-1} \), it is clear that \( r = \frac{1}{3} \).

Step 3: Determine Convergence or Divergence

A geometric series is convergent if the absolute value of the common ratio \( |r| < 1 \). Here, \( |\frac{1}{3}| = \frac{1}{3} < 1 \), so the series converges.

Step 4: Calculate the Sum of the Series

For a convergent geometric series \( \sum_{k=1}^{\infty} ar^{k-1} \), the sum \( S \) is given by the formula \( S = \frac{a}{1 - r} \). Substitute \( a = 4i \) and \( r = \frac{1}{3} \) into the formula to find the sum:\[S = \frac{4i}{1 - \frac{1}{3}} = \frac{4i}{\frac{2}{3}} = 4i \cdot \frac{3}{2} = 6i.\]

Key Concepts

Geometric Series

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

• \( a + ar + ar^2 + ar^3 + \ldots \)

where \( a \) is the first term, and \( r \) is the common ratio. The beauty of geometric series lies in the fact that you can easily determine their convergence. If the absolute value of the common ratio \( |r| < 1 \), the series is convergent and has a finite sum. This sum can be calculated using the formula:

• \( S = \frac{a}{1 - r} \)

For instance, in our example, the first term \( a = 4i \) and the common ratio \( r = \frac{1}{3} \). Since \( |\frac{1}{3}| < 1 \), the series converges to \( S = 6i \). This result shows the ability of geometric series to neatly compress infinite terms into a finite sum.

Complex Numbers

Complex numbers are numbers that have two components: a real part and an imaginary part. They can be expressed as \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part. The letter \( i \) is the imaginary unit, defined as \( i^2 = -1 \).

Complex numbers are crucial in various fields, from engineering to physics, because they allow for solutions to equations that don't have real solutions. Using complex numbers, we can work with terms expressed in complex form in series such as the example: \( \sum_{k=1}^{\infty} 4i(\frac{1}{3})^{k-1} \).

The first term of this series, 4i, highlights the importance of understanding complex numbers, as it represents the series' starting point in complex number form. Moreover, calculations, such as estimating sums of a series, often involve operations on these numbers, enriching the mathematical toolbox.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. In mathematics, determining the behavior of an infinite series—whether it converges to a finite number or diverges—is essential.

For geometric series, as earlier described, convergence depends on the nature of the common ratio. But in more general terms, determining the convergence of an infinite series could involve a variety of tests, each depending on the characteristics of the series involved.

The series in question: \( \sum_{k=1}^{\infty} 4i(\frac{1}{3})^{k-1} \), is a special type of infinite series called a geometric series. Here, the convergence was determined solely by checking if the absolute value of its common ratio was less than one.

Infinite series often pose challenges and opportunities, especially when dealing with series involving complex numbers or those defined by non-standard sequences.