Problem 20 - A First Course in Complex Analysis with Applications

Question

Determine whether the given geometric series is convergent or divergent. If convergent, find its sum. $$ \sum_{k=2}^{\infty} \frac{i^{k}}{(1+i)^{k-1}} $$

Step-by-Step Solution

Answer

The series is convergent and its sum is -1.

Step 1: Identify the First Term

The given geometric series can be expressed in the general form as \( \sum_{k=0}^{\infty} ar^k \), where \( a \) is the first term and \( r \) is the common ratio. First, we need to find the first term. When \( k = 2 \), the first term is \( \frac{i^2}{(1+i)^{2-1}} = \frac{-1}{1+i} \). Thus, \( a = \frac{-1}{1+i} \).

Step 2: Determine the Common Ratio

Next, we identify the common ratio \( r \) of the series. From the series, \( r = \frac{i}{1+i} \). This is obtained by taking the ratio of successive terms: \( \frac{\frac{i^{k+1}}{(1+i)^{k}}}{\frac{i^k}{(1+i)^{k-1}}} = \frac{i}{1+i} \).

Step 3: Check Convergence of the Series

A geometric series converges if the absolute value of the common ratio \( r \) is less than 1, i.e., \( \left| r \right| < 1 \). We compute \( \left| \frac{i}{1+i} \right| = \frac{1}{\sqrt{2}} \) since \( |i| = 1 \) and \( |1+i| = \sqrt{2} \). Since \( \frac{1}{\sqrt{2}} < 1 \), the series is convergent.

Step 4: Compute the Sum of the Series

A convergent geometric series has the sum \( S = \frac{a}{1 - r} \). Substituting our values, \( a = \frac{-1}{1+i} \) and \( r = \frac{i}{1+i} \), the sum \( S \) is computed as \[ S = \frac{\frac{-1}{1+i}}{1 - \frac{i}{1+i}}. \] Simplifying further leads to \[ S = \frac{-1}{1+i - i} = \frac{-1}{1} = -1. \] Thus, the sum of the series is \( -1 \).

Key Concepts

Geometric Series Convergence

When we talk about geometric series convergence, we're dealing with series that have a constant ratio between successive terms. Our goal is to determine whether the series converges, meaning it approaches a specific value, or diverges, meaning it grows without bound. The key factor here is the common ratio, denoted as \( r \). For a geometric series to converge, the absolute value of \( r \) must be less than 1.
This criterion \( |r| < 1 \) ensures that the terms become smaller and smaller as the series progresses. If \( |r| \geq 1 \), the series is divergent.
Understanding convergence is crucial especially when dealing with infinite series since it helps determine if there is a specific sum the series will approach.

Complex Numbers

Complex numbers are numbers that have both real and imaginary parts. The imaginary unit, denoted as \( i \), is defined with the property \( i^2 = -1 \). This makes complex numbers uniquely powerful for calculations involving rotations and oscillations.
When working with series involving complex numbers, it is important to handle both the real and imaginary components. For example, in the original exercise, we deal with terms like \( \frac{i^k}{(1+i)^{k-1}} \).
Breaking down the imaginary components helps ensure that arithmetic remains accurate, and series convergence conditions (like finding absolute values) are met correctly. Always pay attention to both the magnitude and phase of complex numbers.

Sum of a Series

The sum of a geometric series can be found using the formula \( S = \frac{a}{1 - r} \) when the series is convergent. This formula applies only if \( |r| < 1 \).
The term \( a \) represents the first term of the series, while \( r \) is the common ratio. Calculating \( S \) gives us the total value that the infinite series approaches. This concept is critical in many fields, such as physics and economics, where series sum interpretation can denote total lifetime value or accumulated displacement.
The simplicity of this formula allows for quick insight into potentially complex series, illustrating the elegance of convergence properties in geometric series.