Problem 17 - A First Course in Complex Analysis with Applications

Question

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

Step-by-Step Solution

Answer

The series is convergent with sum \( -\frac{1}{5} + \frac{2}{5}i \).

Step 1: Identify the series

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

Step 2: Determine the convergence condition

A geometric series \( \sum a r^k \) is convergent if \(|r| < 1\). Here, \(r = \frac{i}{2}\) and \(|\frac{i}{2}| = \frac{1}{2} < 1\). Therefore, the series is convergent.

Step 3: Use the formula for the sum of a convergent geometric series

The sum \( S \) of a convergent geometric series is given by \( S = \frac{a}{1 - r} \), where \( a \) is the first term. Here, \( a = \frac{i}{2} \) and \( r = \frac{i}{2} \).

Step 4: Calculate the sum using the formula

Substitute \( a = \frac{i}{2} \) and \( r = \frac{i}{2} \) into the formula: \[ S = \frac{\frac{i}{2}}{1 - \frac{i}{2}} \]Simplify the denominator: \[ S = \frac{\frac{i}{2}}{\frac{2-i}{2}} = \frac{i}{2-i} \] Multiply by the conjugate of the denominator to simplify:\[ S = \frac{i}{2-i} \cdot \frac{2+i}{2+i} = \frac{(i)(2+i)}{(2-i)(2+i)} \]\[ S = \frac{2i + i^2}{4+1} = \frac{2i - 1}{5} = \frac{-1}{5} + \frac{2}{5}i \]

Step 5: Simplify the final result

The calculation gives a total sum of \( S = -\frac{1}{5} + \frac{2}{5}i \).

Key Concepts

Convergence of Series

When dealing with series, particularly geometric ones, understanding whether they converge or diverge is crucial. A series converges if the sum of its infinite terms approaches a finite value; otherwise, it diverges. For a geometric series specifically, the convergence depends on the absolute value of its common ratio, denoted as \( r \).

• If \( |r| < 1 \), the series converges.
• If \( |r| \geq 1 \), the series diverges.

In our specific problem, we were given the series \( \sum_{k=1}^{\infty} \left(\frac{i}{2}\right)^{k} \). The common ratio here is \( r = \frac{i}{2} \), which is a complex number. Calculating its magnitude, we find \( |\frac{i}{2}| = \frac{1}{2} \), clearly less than 1.
Thus, this geometric series converges, meaning we can determine a definite sum for all the infinite terms.

Sum of Geometric Series

Once we've established a geometric series converges, the next step is to find its sum. The sum of a convergent infinite geometric series is given by the formula: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term of the series, and \( r \) is the common ratio.
In our geometric series \( \sum_{k=1}^{\infty} \left(\frac{i}{2}\right)^{k} \), both \( a \) and \( r \) are \( \frac{i}{2} \). Plugging these values into the formula:`
\[ S = \frac{\frac{i}{2}}{1 - \frac{i}{2}} \] To solve further, simplifying gives us: \[ S = \frac{i}{2-i} \] To continue simplifying, multiply the numerator and the denominator by the conjugate \( 2+i \) to obtain:\[ S = \frac{(i)(2+i)}{(2-i)(2+i)} = \frac{2i + i^2}{4 + 1} = \frac{2i - 1}{5} = -\frac{1}{5} + \frac{2}{5}i \]
Thus, the sum of the series is \( -\frac{1}{5} + \frac{2}{5}i \).

Complex Numbers in Series

Complex numbers can appear in various mathematical contexts, including series. Understanding how to handle them is essential for solving problems involving geometric series with complex ratios.
A complex number is expressed as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. It's important to recognize that operations with complex numbers follow different rules than those strictly involving real numbers.

• The absolute value or magnitude of a complex number \( z = a + bi \) is calculated as \( |z| = \sqrt{a^2 + b^2} \).
• When simplifying expressions with complex numbers, the conjugate \( a - bi \) is often used to rationalize denominators.

In our geometric series example, the common ratio \( \frac{i}{2} \) is a complex number. Its magnitude is \( |\frac{i}{2}| = \frac{1}{2} \), pivotal in determining the convergence of the series.
Handling the sum's computation involved using the conjugate of the denominator to simplify a fraction, highlighting the significance of complex numbers' properties in series calculations.