Problem 19 - Advanced Engineering Mathematics

Question

In Problems 15-20, determine whether the given geometric series is convergent or divergent. If convergent, find its sum. $$ \sum_{k=0}^{\infty} 3\left(\frac{2}{1+2 i}\right)^{k} $$

Step-by-Step Solution

Answer

The series is convergent and its sum is -3.

Step 1: Identify the Geometric Series

The given series is \( \sum_{k=0}^{\infty} 3 \left( \frac{2}{1+2i} \right)^k \). This is a geometric series with the first term \( a = 3 \) and the common ratio \( r = \frac{2}{1+2i} \).

Step 2: Determine Convergence Criteria

A geometric series \( \sum_{k=0}^{\infty} ar^k \) converges if and only if the absolute value of the common ratio \( |r| < 1 \).

Step 3: Calculate Absolute Value of r

Calculate \( |r| = \left| \frac{2}{1+2i} \right| \). To find this, compute \( \left| \frac{2}{1+2i} \right| = \frac{|2|}{|1+2i|} = \frac{2}{\sqrt{1^2 + (2)^2}} = \frac{2}{\sqrt{5}} \).

Step 4: Check Convergence

Since \( |r| = \frac{2}{\sqrt{5}} < 1 \), the series is convergent.

Step 5: Calculate the Sum of the Series

The sum of a convergent geometric series \( \sum_{k=0}^{\infty} ar^k \) is given by \( \frac{a}{1-r} \). So, the sum is \[ \frac{3}{1 - \frac{2}{1+2i}} \].

Step 6: Simplify Sum Formula

Simplify the expression \( \frac{3}{1 - \frac{2}{1+2i}} \):\[ 1 - \frac{2}{1+2i} = \frac{1+2i}{1+2i} - \frac{2}{1+2i} = \frac{1+2i - 2}{1+2i} = \frac{-1+2i}{1+2i} \]Hence, the sum is \[ \frac{3}{\frac{-1+2i}{1+2i}} = 3 \cdot \frac{1+2i}{-1+2i} = \frac{3(1+2i)}{-1+2i} \].

Step 7: Further Simplification Using Complex Conjugates

Multiply numerator and denominator by the conjugate of the denominator:\[ \frac{3(1+2i)}{-1+2i} = \frac{3(1+2i)(-1-2i)}{(-1+2i)(-1-2i)} \]This leads to \[ \frac{3(-1-2i + 2i + 4i^2)}{5} \] simplifying down to \[ \frac{3(-5)}{5} = -3 \].

Key Concepts

Complex Numbers

Complex numbers extend our familiar number system and are represented in the form \( a + bi \) where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, satisfying \( i^2 = -1 \). In the problem above, we encounter complex numbers when we see terms like \( 1 + 2i \). These numbers are crucial in advanced mathematics because they allow for solutions beyond the realm of real numbers.

Complex numbers can be added, subtracted, multiplied, and divided similarly to real numbers. However, while performing division, the complex conjugate plays an essential role. The conjugate of a complex number \( a + bi \) is \( a - bi \). Multiplying both the numerator and denominator of our expression by the conjugate helps to simplify the division by eliminating the imaginary part in the denominator. This was applied in the step where the term \(-1 + 2i\) was divided in the formula and simplified with its conjugate \(-1 - 2i\).

Understanding how to manipulate these numbers is key to solving many problems involving geometric series with complex terms.

Absolute Value

In the context of complex numbers, absolute value refers to the magnitude or size of the number. For a complex number \( z = a + bi \), the absolute value is given by \( |z| = \sqrt{a^2 + b^2} \). This corresponds to the distance from the origin to the point \( (a, b) \) in the complex plane.

Within the exercise, the absolute value was used to determine the convergence of the series by examining \( |r| \), where \( r = \frac{2}{1+2i} \). Calculating the absolute value helps to understand whether this series converges or not by giving us a precise measure of the ratio's size.

Solving for \( |r| \) included:

• Finding the absolute values of the numerator and the denominator separately.
• Combining them using division.

This process highlights the importance of understanding both the definition of absolute value and how it operates within complex arithmetic.

Sum of a Series

A geometric series consists of terms that have a constant ratio between successive members, expressed as \( ar^k \) in the infinite series \( \sum_{k=0}^{\infty} ar^k \). If this series converges, its sum can be found using the formula \( \frac{a}{1-r} \) when the absolute value of the common ratio \( |r| < 1 \).

In our specific problem, where \( a = 3 \) and \( r = \frac{2}{1+2i} \), we found the sum after confirming convergence. Understanding how to express and simplify the terms in the sum formula, especially when involving complex numbers, was key.

The steps to find the sum included:

• Substituting \( r \) into the sum formula.
• Simplifying the resulting complex expression.

This requires using complex arithmetic cleverly, such as conjugates, to simplify to a real number result.

Convergence Criteria

To determine whether an infinite series converges, certain criteria need to be verified. For a geometric series \( \sum_{k=0}^{\infty} ar^k \), convergence hinges on the magnitude of the common ratio \( r \).

The specific criterion is that the absolute value \( |r| \) must be less than 1. If it holds true, the series converges; otherwise, it diverges. In the exercise, calculating \( |r| = \frac{2}{\sqrt{5}} \) established that the series is convergent as \( \frac{2}{\sqrt{5}} < 1 \).

This also demonstrates the importance of understanding absolute values in the context of complex numbers. The convergence criterion allows us to conclude not only if a series converges, but also opens the door to finding its exact sum, thereby solving the problem entirely.