Question

Von einer Potenzreihe \(\sum_{k=0}^{\infty} a_{k}(x-2)^{k}, a_{k} \in \mathbb{R}\), weiß man, dass sie für \(x=5\) absolut konvergent ist, für \(x=-2\) konvergent und für \(x=-4\) divergent ist. Was kann man über den Konvergenzradius aussagen? In welchen Intervallen liegt mit Sicherheit Konvergenz bzw. Divergenz vor?

Short answer

The convergence radius is \(3 < R \leq 4\). The series converges in \(x \in (-2, 6)\) and diverges for \(x \leq -4\).

Step-by-step solution

Understanding Power Series Convergence

The power series \(\sum_{k=0}^{\infty} a_{k}(x-2)^{k}\) is centered at 2. A power series \(\sum a_k (x-c)^k\) has a radius of convergence \(R\), which determines the interval \(x \in (c-R, c+R)\) where the series converges.

Utilize Known Convergence Information

We know that the series converges absolutely at \(x = 5\). This implies \(|5-2| < R\), meaning \(R > 3\). There is convergence at \(x = -2\) so \(|-2-2| = 4 \leq R\). The series diverges at \(x = -4\), so \(|-4-2| = 6 > R\).

Determine the Convergence Radius

From the given informations, the radius \(R\) must satisfy \(3 < R \leq 4\). Consequently, this range fits all the criteria of absolute convergence, general convergence, and divergence provided.

Find Intervals of Guaranteed Convergence and Divergence

With \(\sum a_k (x-2)^k\) having a radius of convergence \(R\) with \(3 < R \leq 4\), the series converges for any \(x\) satisfying \(2-R < x < 2+R\). As \(x = -4\) diverges, values \(x \leq -4\) guarantee divergence.

Key concepts

Understanding the Radius of Convergence

The radius of convergence, often denoted by \( R \), is a crucial concept in understanding power series. For a power series centered at \( c \), represented as \( \sum_{k=0}^{\infty} a_k (x-c)^k \), \( R \) determines the range of \( x \) values for which the series converges. This range is defined as \( x \in (c-R, c+R) \).
The determination of \( R \) utilizes known values where the series is absolutely convergent. For instance, if the series converges absolutely at \( x = 5 \), we calculate the distance from the center \( c \) (which is 2 in our exercise): \(|5 - 2| < R\), leading to \( R > 3 \). However, the series is known to diverge at \( x = -4 \), thus \(|-4 - 2| = 6 > R\).
By combining these inequalities, we conclude \( 3 < R \leq 4 \). Thus, the radius of convergence encompasses all values of \( x \) that satisfy this relationship.

Exploring Absolute Convergence

Absolute convergence is a stronger form of convergence that implies convergence but not vice versa. For a power series \( \sum a_k (x-c)^k \), absolute convergence at a particular value \( x \) means the series \( \sum |a_k (x-c)^k| \) also converges.
This concept is crucial because absolute convergence provides stronger evidence regarding the behavior of the series. In our exercise, knowing that the series converges absolutely at \( x = 5 \) ensures that relative to the center \( x = 2 \), the absolute distance \(|5-2| = 3\) is less than \( R \). This fact was essential in helping us determine that \( R > 3 \).

Identifying the Convergence Interval

The convergence interval is the range of \( x \) values for which the power series converges. This interval is tightly linked to the radius of convergence \( R \).
By understanding the radius from our exercise as \( 3 < R \leq 4 \), the convergence interval, based on the center \( c = 2 \), is \((2-R, 2+R)\). So, this means the series converges strictly within the domain \(2-4 < x < 2+4\) which simplifies to \(-2 < x < 6\).
Within this range, every value of \( x \) ensures that the series converges, fulfilling its mathematical definition and making sure no divergence happens.

Criteria for Divergence

Divergence in a power series occurs when the series does not add up to a finite value or does not maintain any form of convergence. Understanding divergence can help define the boundaries of convergence.
For instance, if the given series diverges at \( x = -4 \), it suggests that beyond these boundaries, specifically for \( x \leq -4 \), the series will not converge. This divergence criteria is directly related to the endpoints of convergence interval.
Divergence criteria are crucial for framing the limits of the series' behavior. If convergence is assured within a calculated interval, the complementary segments provide confidence in divergence for those points where the radius no longer encapsulates the convergence range.