Chapter 3: Problem 25
With a short argument using what you know about the harmonic series, show that the \(p\) -harmonic series for \(0
Short Answer
Expert verified
The \( p \)-harmonic series diverges for \( 0 < p \leq 1 \) because it exceeds the divergent harmonic series terms.
Step by step solution
01
Understand the Harmonic Series
The harmonic series is the sum of reciprocals of all positive integers: \( \sum_{n=1}^{\infty} \frac{1}{n} \). This series is known to be divergent.
02
Define the p-Harmonic Series
The \( p \)-harmonic series is defined as the series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). For \( p=1 \), this is simply the standard harmonic series, which is divergent.
03
Explore the Case of 0 < p < 1
For \( 0 < p < 1 \), each term of the series \( \frac{1}{n^p} > \frac{1}{n} \) because raising \( n \) to a smaller power will result in a larger quantity. Thus, each term of the \( p \)-harmonic series is greater than each corresponding term in the original harmonic series.
04
Apply the Comparison Test
Since each term of the \( p \)-harmonic series is greater than each corresponding term of the divergent harmonic series, the \( p \)-harmonic series 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.
Harmonic Series
The Harmonic Series is a fundamental concept in the study of infinite series. It consists of the sum of the reciprocals of all positive integers, expressed as: \[ \sum_{n=1}^{\infty} \frac{1}{n} \]. This series is particularly interesting because, despite its terms getting smaller and smaller, the overall sum keeps increasing without bound. Hence, the Harmonic Series is known to be divergent.
- "Divergent" means the series does not settle down to a finite sum.
- The divergence of the Harmonic Series is a classic example in calculus since it looks like it might converge at first glance due to the decreasing terms.
- To intuitively understand divergence here, imagine adding infinitely small positive numbers that, though they shrink, are endless.
p-Harmonic Series
Building on the Harmonic Series, we encounter the concept of the \( p \)-Harmonic Series. This series extends the idea of the Harmonic Series by introducing an exponent \( p \), forming the series:\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]. The interesting behavior of this series depends heavily on the value of \( p \).
- When \( p = 1 \), it matches the Harmonic Series, known to be divergent.
- For \( 0 < p < 1 \), each term \( \frac{1}{n^p} \) is actually larger than \( \frac{1}{n} \), since raising \( n \) to a power less than one increases the value of the denominator less than when \( n \) is raised to the power of one.
- This results in a series with terms that are even larger than those in the Standard Harmonic Series for \( 0 < p < 1 \).
Comparison Test
The Comparison Test is a fundamental technique in determining the convergence or divergence of a series by comparing it to another series with known behavior.
This test states that if you have two series \( \sum a_n \) and \( \sum b_n \), where \( a_n \leq b_n \) for all \( n \) sufficiently large, and if the series \( \sum b_n \) is known to be convergent, then the series \( \sum a_n \) must also converge.
This test states that if you have two series \( \sum a_n \) and \( \sum b_n \), where \( a_n \leq b_n \) for all \( n \) sufficiently large, and if the series \( \sum b_n \) is known to be convergent, then the series \( \sum a_n \) must also converge.
- Conversely, if \( a_n \geq b_n \) for all \( n \) large enough, and \( \sum b_n \) diverges, then \( \sum a_n \) will also diverge.
- This works because, intuitively, a larger term than a divergent series' term cannot sum to a finite number.
- In the case of the \( p \)-Harmonic Series for \( 0 < p < 1 \), since each \( \frac{1}{n^p} \) is greater than \( \frac{1}{n} \), the series diverges like the harmonic series.