Problem 63

Question

Assume that the series $\sum a_{n}(x-2)^{n}$ converges for $x=-1$ and diverges for $x=6 .$ Answer true $(\mathrm{T})$ , false $(\mathrm{F})$ , or not enough information given $(\mathrm{N})$ for the following statements about the series. $$ \begin{array}{l}{\text { a. Converges absolutely for } x=1} \\ {\text { b. Diverges for } x=-6} \\ {\text { c. Diverges for } x=2} \\ {\text { d. Converges for } x=0} \\ {\text { e. Converges absolutely for } x=5} \\\ {\text { f. Diverges for } x=4.9} \\ {\text { g. Diverges for } x=5.1} \\\ {\text { a. Diverges for } x=5.1} \\ {\text { h. Converges absolutely for } x=4}\end{array} $$

Step-by-Step Solution

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Answer

a: True, b: True, c: False, d: True, e: False, f: False, g: True, h: False.

1Step 1: Identify the Convergence Information

To solve the problem, we need to understand the convergence properties of the series $ \sum a_n (x-2)^n $. It's given that the series converges when $x = -1$ and diverges when $x = 6$. This indicates that the interval of convergence is between these two values with the radius of convergence $ R $ around the center $ x = 2 $. This implies $ |-1 - 2| = 3 $ for convergence at $ x = -1 $ and $ |6 - 2| = 4 $ for divergence at $ x = 6 $. Hence, the center is 2 with radius 3, having $-1\leq x \leq 5 $.

2Step 2: Evaluate Statements About Convergence

a. For $ x = 1 $, check $|x - 2| < R$: |-1| = 1 < 3. **True, converges absolutely.**b. For $ x = -6 $, check $ |-6 - 2| = 8$. **True, diverges as it is outside interval.**c. For $ x = 2 $, the series is centered at 2. **False, this is the center, series convergent.** d. For $ x = 0 $, check $|0-2| = 2 < 3$. **True, converges.** e. For $ x = 5 $, check $|5-2| = 3$: **False, it converges but not absolutely. (At boundary of convergence)** f. For $ x = 4.9 $, check $|4.9-2| = 2.9 < 3 $. **False, converges not diverges as it is in interval.** g. For $ x = 5.1 $, check $|5.1-2| = 3.1 > 3$. **True, diverges as it is outside interval.** h. For $ x = 4 $, check \(|4-2|=2

Key Concepts

Radius of ConvergenceAbsolute ConvergenceDivergenceInterval of Convergence

Radius of Convergence

The radius of convergence is a fundamental concept in understanding the behavior of a power series, such as $\sum a_n (x-2)^n$. It helps us determine how far from the center (in this case, 2) the series will still converge. This distance is called the radius of convergence, represented by $ R $.

In the given exercise, the series converges for $ x = -1 $, and diverges for $ x = 6 $. This tells us the boundary of convergence is closer at -1 and farther at 6.

The distance from 2 to -1 is $ |-1 - 2| = 3 $.
The distance from 2 to 6 is $ |6 - 2| = 4 $, indicating the divergence.

When assessing convergence using the radius, always check whether the distance $|x-2|$ is less than $ R = 3 $. If it is less, the series will converge at that point. Knowing $ R $ is crucial for determining the series' behavior within an interval centered around 2.

Absolute Convergence

Absolute convergence is when a series converges not just for the terms themselves, but also when all terms are made positive by taking absolute values. This form of convergence is stronger than regular convergence. If a series converges absolutely, it also converges normally.

Consider the series in the exercise:If you want to check absolute convergence for a particular value, say $ x = 1 $, you calculate $|x-2|$ and compare with $ R $ = 3:

$|1 - 2| = 1$ is less than 3, suggesting convergence.Absolute convergence often requires checking whether the series of $|a_n|(x-2)^n$ converges.
For instance, at a boundary like $ x = 5 $ ($|5 - 2| = 3$), it converges but not absolutely since it lies right on the edge.

Keep in mind, absolute convergence at a point implies strong convergence is present, ensuring stability within that region.

Divergence

Divergence occurs when a series does not settle on a finite limit, meaning it grows beyond bounds as more terms are added. In simple terms, it means getting bigger and bigger without stopping.

In the present scenario:The series diverges for $ x = 6 $. This helps set a frame for understanding the limits:

If $|x-2|$ becomes greater than 3 ($ R = 3 $), divergence will happen.
For example, at $ x = 5.1 $ where $|5.1 - 2| = 3.1$ exceeds 3, divergence will be expected.

Divergence checks are essential when determining the edges of a series' behavior, ensuring comprehension of where stability drops.

Interval of Convergence

The interval of convergence defines the range of $ x $ values for which a series converges. Within this range, every point satisfies convergence conditions.

For the exercise's series around center 2:The given points establish $-1 \le x \le 5$ as the interval. This is found using:

Given convergence at $ x = -1 $ establishing one boundary through $ |-1 - 2| = 3 $.
Checking against divergence at $ x = 6 $ verifies limits across boundaries for $|6 - 2|$.

Within this range:

Any $ x $ complying with $|x - 2| < 3$fits inside the interval for regular convergence.
Close margins, such as those near boundaries like $ x = 5 $, warrant careful consideration for energies of absolute versus regular limits.

These concepts ensure understanding of the regions with finite responses against growing sizes.

Problem 63

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