Chapter 9: Problem 13
Determine whether the series converges. $$ \sum_{k=1}^{\infty} \frac{1}{\sqrt[3]{2 k-1}} $$
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Recognize the Series Type
The given series is \( \sum_{k=1}^{\infty} \frac{1}{\sqrt[3]{2k-1}} \). It is an infinite series of the form \( \sum_{k=1}^{\infty} a_k \), where \( a_k = \frac{1}{(2k-1)^{1/3}} \).
02
Identify the Comparison Test
We will use the Comparison Test to determine the convergence of the series. Find a simpler series, \( \sum b_k \), to compare with \( \sum a_k \). A natural choice for \( b_k \) is \( \frac{1}{k^{1/3}} \), which resembles a known \( p \)-series.
03
Analyze the Comparison Series
The series \( \sum_{k=1}^{\infty} \frac{1}{k^{1/3}} \) is a \( p \)-series with \( p = \frac{1}{3} \). A \( p \)-series \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). Since \( \frac{1}{3} \leq 1 \), this series diverges.
04
Establish the Comparison
We note that since \( 2k - 1 \geq k \) for all \( k \geq 1 \), it follows that \( \frac{1}{(2k-1)^{1/3}} \geq \frac{1}{k^{1/3}} \). Hence, \( a_k \geq b_k \) for all \( k \geq 1 \).
05
Apply the Comparison Test
Since \( \sum_{k=1}^{\infty} b_k = \sum_{k=1}^{\infty} \frac{1}{k^{1/3}} \) diverges and \( a_k \geq b_k \) for all \( k \), the series \( \sum_{k=1}^{\infty} a_k = \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{1/3}} \) must also diverge by the Comparison Test.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
comparison test
The Comparison Test is a handy tool for determining the convergence of an infinite series. It involves comparing a complex series to a simpler, well-known one. If you can establish a clear link between the two, you can use the behavior of the known series to understand the series in question. Here's how it works:
- Identify two series: the one you're examining (\( \sum a_k \)), and a simpler one (\( \sum b_k \)) to compare it to.
- If \( a_k \leq b_k \) for all \( k \), and the simpler series \( \sum b_k \) converges, then \( \sum a_k \) also converges.
- If \( a_k \geq b_k \) for all \( k \), and \( \sum b_k \) diverges, then \( \sum a_k \) also diverges.
- Judging the inequality between series terms depending on convergence or divergence.
- A thorough understanding of how the simpler series behaves.
p-series
A p-series is a fundamental type of series that looks like this: \( \sum_{k=1}^{\infty} \frac{1}{k^p} \). It's called a "p-series" because the behavior of the series strongly depends on the value of \( p \):- If \( p > 1 \), the series converges.- If \( p \leq 1 \), the series diverges.This makes p-series incredibly straightforward to analyze since checking convergence or divergence is just a matter of looking at \( p \).In our example, when trying to compare it with the known p-series \( \sum_{k=1}^{\infty} \frac{1}{k^{1/3}} \):- Here, \( p = \frac{1}{3} \), which is less than 1.- So, this p-series diverges.Knowing how a p-series behaves allows us to quickly determine the behavior of more complicated series when we can align them with a p-series in a comparison.
infinite series
An infinite series is essentially the sum of an infinite sequence of terms. Mathematically, it is represented as \( \sum_{k=1}^{\infty} a_k \), where each \( a_k \) is a term in the sequence. The key question for any infinite series is whether it converges (sum approaches a finite number) or diverges (sum grows without bound).To determine convergence:
- Various tests, like the Comparison Test, can be used.
- Each test depends on specific criteria related to the terms of the series.