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Chapter 9: Problem 14

Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4 / 3}} $$

Short Answer

Expert verified
The series is absolutely convergent.

Step by step solution

01

Identify the Series Type

The given series is \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4/3}} \). This is an alternating series because of the factor \((-1)^{k+1}\), which causes the signs of the terms to alternate.
02

Test for Absolute Convergence

To check for absolute convergence, we consider the series of the absolute values of its terms: \( \sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k^{4/3}} \right| = \sum_{k=1}^{\infty} \frac{1}{k^{4/3}} \). Note that this is a p-series with \( p = \frac{4}{3} > 1 \). It is known that a p-series converges if \( p > 1 \). Therefore, this series is absolutely convergent.
03

Conclusion

Since the series \( \sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k^{4/3}} \right| \) converges, the original alternating series \( \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4/3}} \) is absolutely convergent by definition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alternating Series
An alternating series is a series where the signs of the terms alternate between positive and negative. The mathematical representation often includes a factor such as \((-1)^k\) or \((-1)^{k+1}\), which is what causes the change in sign. In the exercise provided, we have the series: \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4/3}} \). This indicates that each term alternates in sign, e.g., the first term is positive, the second is negative, and so on.
  • Alternating series can converge even if the non-alternating series (i.e., ignoring the \((-1)^k\) factor) diverges.
  • To determine if an alternating series converges, we often use the Alternating Series Test. This test states that an alternating series \(\sum (-1)^k a_k\) converges if:
    • The absolute value of the sequence \(a_k\) is monotonically decreasing, and
    • \(\lim_{k \to \infty} a_k = 0\).
Absolute Convergence
Absolute convergence is a stronger form of convergence meant to ensure that a series is indeed stable regardless of the sign of its terms. A series \(\sum a_k\) is considered absolutely convergent if the series of absolute values \(\sum |a_k|\) converges.

In the original exercise, we encounter absolute convergence when we examine: \(\sum_{k=1}^{\infty} \left| \frac{(-1)^{k+1}}{k^{4/3}} \right| = \sum_{k=1}^{\infty} \frac{1}{k^{4/3}} \).
  • If a series is absolutely convergent, it implies that the original series converges as well (this is known as conditional convergence).
  • Absolute convergence is generally easier to handle since the alternating nature of the terms isn't considered in determining convergence.
  • Also, series that are absolutely convergent are generally more stable and are easier to rearrange while still maintaining convergence.
P-Series
A p-series is a specific type of series given by \(\sum_{k=1}^{\infty}\frac{1}{k^p}\). The convergence of a p-series is determined by the value of \(p\).
  • If \(p > 1\), the p-series converges. This is a crucial threshold for ensuring p-series stability, and it comes in handy while checking absolute convergence in other series.
  • If \(p \leq 1\), the series diverges, meaning the sum doesn't settle to any finite limit.
In our given example, the absolute series \(\sum_{k=1}^{\infty} \frac{1}{k^{4/3}}\) is a p-series with \({p = \frac{4}{3}} > 1\). Therefore, this series is guaranteed to converge.

Understanding p-series helps students recognize functions quickly, especially in determining convergence patterns without delving into deep calculations.

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Most popular questions from this chapter

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