Problem 29 - Advanced Engineering Mathematics

Question

Show that the power series \(\sum_{k=1}^{\infty} \frac{(z-i)^{k}}{k 2^{k}}\) is not absolutely convergent on its circle of convergence. Determine at least one point on the circle of convergence at which the power series converges.

Step-by-Step Solution

Answer

The series is not absolutely convergent on its circle of convergence. It converges at a point like \( z = 2+i \) on the circle.

Step 1: Determine the Radius of Convergence

To find the radius of convergence of the power series \( \sum_{k=1}^{\infty} \frac{(z-i)^{k}}{k \cdot 2^{k}} \), use the formula \( \frac{1}{R} = \limsup_{k \to \infty} |a_k|^{1/k} \) where \( a_k = \frac{1}{k \cdot 2^k} \). Here \( |a_k|^{1/k} = \left( \frac{1}{k \cdot 2^k} \right)^{1/k} = \frac{1}{k^{1/k} \cdot 2} \). As \( k \to \infty \), \( k^{1/k} \to 1 \), so \( \limsup_{k \to \infty} \frac{1}{k^{1/k} \cdot 2} = \frac{1}{2} \). Thus the radius of convergence \( R = 2 \).

Step 2: Analyze Absolute Convergence on the Circle

Check absolute convergence at \( |z-i| = 2 \). The series is \( \sum_{k=1}^{\infty} \frac{|z-i|^{k}}{k \cdot 2^{k}} \). At the boundary, \( |z-i| = 2 \), the terms become \( \frac{2^{k}}{k \cdot 2^{k}} = \frac{1}{k} \). The resulting series \( \sum_{k=1}^{\infty} \frac{1}{k} \) is the harmonic series, which diverges. Therefore, the original series is not absolutely convergent on the circle of convergence.

Step 3: Determine Point of Convergence on the Circle

To find a point where the series converges, consider \( z = 1+i \). Here, \( |1+i - i| = |1| = 1 < 2 \), within the radius of convergence, thus the series converges normally. However, on the circle itself for restricted series, choose \( z = 2+i \), so \( |2+i - i| = |2| = 2 \). For real parts affecting alternating series strategies, note \( z-i = 2 \) giving a series \( \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \), which converges by the Alternating Series Test.

Key Concepts

Radius of Convergence

When working with power series, understanding the radius of convergence is crucial. The radius of convergence tells us the extent of values over which the series converges to a sum. For the series \( \sum_{k=1}^{\infty} \frac{(z-i)^{k}}{k \cdot 2^{k}} \), the formula used to find this radius is:
\[\frac{1}{R} = \limsup_{k \to \infty} |a_k|^{1/k}\]where \( a_k = \frac{1}{k \cdot 2^k} \).

In this case, as \( k \) tends to infinity, the term \( k^{1/k} \) approaches 1. Therefore, the value \( \frac{1}{k^{1/k} \cdot 2} \) simplifies to \( \frac{1}{2} \), which tells us that the radius \( R \) is 2.
This radius depicts a circle centered at \( z = i \) with a radius of 2.

When inside this circle, the series converges. However, what happens at or outside this circle needs further investigation.

Absolute Convergence

Absolute convergence involves looking at the absolute values of the series terms and determining whether the modified series converges. For the power series in question, examining absolute convergence at its circle of convergence \(|z-i| = 2\) is necessary.

We consider the series:
\[\sum_{k=1}^{\infty} \frac{|z-i|^{k}}{k \cdot 2^{k}}\]At \(|z-i| = 2\), the terms become \( \frac{2^k}{k \cdot 2^k} = \frac{1}{k} \), which is the harmonic series.
The harmonic series \( \sum_{k=1}^{\infty} \frac{1}{k} \) is known to diverge.

This indicates that the power series does not converge absolutely on its circle of convergence.

• Absolute convergence ensures more stability in a series' sum.
• If a series converges absolutely, it also converges normally.
• However, diverging in absolute terms only restricts this assurance.

Alternating Series Test

To analyze specific points on the circle of convergence where the series might converge, the Alternating Series Test becomes helpful. This test provides conditions under which series with alternating signs converge.

Consider the point \( z = 2+i \) on the circle where \(|z-i| = 2\). Here, \( z-i = 2 \) leads to the series:
\[\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\]This series alternates in sign. For the Alternating Series Test to confirm convergence, two conditions must be met:

• The absolute value of successive terms \(|a_k|\) decreases monotonically.
• The limit of \( a_k \) as \( k \to \infty \) is zero.

Both conditions are satisfied here:

• \( |a_k| = \frac{1}{k} \) does indeed decrease.
• As \( k \) approaches infinity, \( a_k \to 0 \).

Thus, by the Alternating Series Test, this specific point allows the series to converge, even though it isn’t absolutely convergent. This makes understanding series behavior at the circle’s edge richer and more nuanced.