Problem 21 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=1}^{\infty}\left(2+(-1)^{n}\right) \cdot(x+1)^{n-1} $$

Step-by-Step Solution

Answer

The series converges absolutely for \(-2 < x < 0\).

Step 1: Analyze Series Function

Observe that the given series is \( \sum_{n=1}^{\infty} (2 + (-1)^n) (x+1)^{n-1} \). The term \((2 + (-1)^n)\) alternates between 1 and 3 depending on whether \(n\) is odd or even.

Step 2: Find Power Series Form

Write the series as a combination of two separate series: \( \sum_{k=0}^{\infty} \left[(3)(x+1)^{2k} + (1)(x+1)^{2k+1}\right] \). This separates the terms based on odd and even \(n\).

Step 3: Ratio Test Preparation

Use the Ratio Test to find the radius of convergence. We're interested in the limit \( \lim_{n \to \infty} \left| \frac{c_{n+1}(x+1)^{n}}{c_n(x+1)^{n-1}} \right| \), where \( c_n = 2 + (-1)^n \).

Step 4: Apply the Ratio Test

Calculate the limit: \( \lim_{n \to \infty} \left| \frac{(2 + (-1)^{n+1})(x+1)}{2 + (-1)^n} \right| = |x+1| \). The series converges when \(|x+1| < 1\). Therefore, the radius of convergence is 1.

Step 5: Determine the Interval of Convergence

Since \(|x+1| < 1\), the interval of convergence is \(-2 < x < 0\). We need to check convergence at the endpoints \(x = -2\) and \(x = 0\).

Step 6: Test Endpoint x = -2

Substitute \(x = -2\) into the original series: \( \sum (2 + (-1)^n)(-1)^{n-1} \). Simplify to find \( \sum (-1)^n \), which diverges, so \(x = -2\) is not included in the interval.

Step 7: Test Endpoint x = 0

Substitute \(x = 0\) into the original series: \( \sum (2 + (-1)^n) \). Simplify to get \(\sum 1 + (-1)^n \), which diverges as the terms do not tend to zero. Hence, \(x = 0\) is also not included.

Step 8: Conclusion on Series Convergence

The series converges absolutely for \(-2 < x < 0\) because within this interval, the ratio test shows convergence. The endpoints do not support convergence under absolute or conditional terms.

Key Concepts

Interval of Convergence

When dealing with power series, finding the interval of convergence is crucial. The interval of convergence includes all the values of the variable for which the power series converges. Understanding this interval is vital for determining the behavior of the series across different values.
In the given series, we're working with the series expression:
\[ \sum_{n=1}^{\infty} (2+(-1)^{n})(x+1)^{n-1} \]
This expression is a blend of alternating terms, hopping between values based on whether the power index is odd or even.
To find the interval of convergence, we use tests like the Ratio Test, which helps us determine when the absolute value of \( x+1 \) is less than the radius of convergence. In this exercise, the Ratio Test tells us that \( |x+1| < 1 \), giving a radius of convergence of 1. Hence, solving for \( x \), we have:

• The interval \( -2 < x < 0 \)

This means the series converges for all \( x \) within this interval, except for the endpoints. Subsequently, testing the endpoints tells us whether they are also included in the interval.

Absolute Convergence

Understanding absolute convergence is key to analyzing series. In essence, a series converges absolutely when the series formed by taking the absolute value of its terms converges. This is a stronger condition than ordinary convergence because absolute convergence often implies the regular convergence of the series itself.
For the given series:
\[ \sum_{n=1}^{\infty} (2+(-1)^{n})(x+1)^{n-1} \]
Within the interval of convergence \( -2 < x < 0 \), the series converges absolutely because this range ensures that the terms \( |x+1| \) are sufficiently small to meet the convergence criteria.
For convergence within this interval, we can apply the Ratio Test to the series of absolute values, confirming that it holds for these \( x \) values.

• Absolute convergence generally implies strong stability, meaning the order of terms in the series does not affect its convergence.

Although we determined that the original series does not absolutely converge at the endpoints \( x = -2 \) and \( x = 0 \), the open interval checked does support absolute convergence.

Conditional Convergence

Conditional convergence is where it gets more subtle. In this concept, a series converges, but it does not do so absolutely. This typically happens with series having both positive and negative terms that partially "cancel out" each other to reach a finite sum.
For the series in question:
\[ \sum_{n=1}^{\infty} (2+(-1)^{n})(x+1)^{n-1} \]
We established that the interval of convergence is \( -2 < x < 0 \). However, within this interval, all convergences are absolute; there is no x-value yielding only conditional convergence. The endpoints, \( x = -2 \) and \( x = 0 \), do not allow for conditional convergence either, as they completely diverge.

• Conditional convergence can often be seen in series that alternate and "balance" themselves around zero.

The absence of conditional convergence in this case doesn't detract from understanding its potential significance, especially in different series where absolute convergence is not achievable.