Chapter 11: Problem 45
Find all positive values of \(b\) for which the series \(\sum_{n=1}^{\infty} b^{\ln n}\) converges.
Short Answer
Expert verified
The series converges for positive values of \( b \) such that \( 0 < b < 1 \).
Step by step solution
01
Understand Series Convergence
The given series is \( \sum_{n=1}^{\infty} b^{\ln n} \). To determine when this series converges, we need to analyze the behavior of the general term \( a_n = b^{\ln n} \). By simplifying, we recognize this as \( a_n = n^{\ln b} \).
02
Apply the Limit Test for Series
For the series \( \sum_{n=1}^{\infty} a_n \) to converge, it is required that \( \lim_{n \to \infty} a_n = 0 \). In our problem, this becomes \( \lim_{n \to \infty} n^{\ln b} = 0 \). Analyzing this limit, note that for any \( b > 1 \), \( \ln b > 0 \) and \( n^{\ln b} \to \infty \), while for \( b < 1 \), \( \ln b < 0 \) and \( n^{\ln b} \to 0 \).
03
Determine the Convergence Condition
Since \( \lim_{n \to \infty} n^{\ln b} = 0 \) only when \( \ln b < 0 \), we require \( b < 1 \) for the terms to approach zero, a necessary condition for convergence. Hence, the series converges only for positive \( b < 1 \).
04
Conclude with the Positive Value Condition
To ensure \( b \) remains positive, the condition \( b < 1 \) and \( b > 0 \) must both be satisfied. Therefore, the allowed positive values of \( b \) for which the series converges are \( 0 < b < 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Test for Series
The Limit Test for Series is a foundational concept in determining whether an infinite series converges or diverges. When dealing with a series like \( \sum_{n=1}^{\infty} a_n \), one of the simplest tests involves the behavior of the general term \( a_n \) as \( n \) approaches infinity. Specifically, if \( \lim_{n \to \infty} a_n eq 0 \), the series must diverge. This is based on the logical premise that if the terms do not increasingly get smaller and approach zero, then the series cannot settle to a finite limit.
In the exercise, the general term is given by \( a_n = n^{\ln b} \). By applying the limit test, we analyze \( \lim_{n \to \infty} n^{\ln b} \). For convergence:
This test helps in quickly assessing the behavior of the series without needing complex transformations or deeper analyses. It's a critical first step in series convergence problems.
In the exercise, the general term is given by \( a_n = n^{\ln b} \). By applying the limit test, we analyze \( \lim_{n \to \infty} n^{\ln b} \). For convergence:
- If \( \ln b > 0 \), \( n^{\ln b} \to \infty \), hence the series diverges.
- If \( \ln b < 0 \), \( n^{\ln b} \to 0 \), fulfilling the convergence criteria.
This test helps in quickly assessing the behavior of the series without needing complex transformations or deeper analyses. It's a critical first step in series convergence problems.
Convergence Condition
The convergence condition is vital for deciding when a series will settle at a finite sum. This condition relies heavily on the limit behavior noted in the Limit Test for Series.
For the series \( \sum_{n=1}^{\infty} n^{\ln b} \), convergence depends on the sign of \( \ln b \):
Therefore, the primary convergence condition is established by requiring \( b < 1 \) for convergence, ensuring \( \ln b \) is negative. This delicate balance of conditions allows us to deduce that the values of \( b \) which result in convergence are those where 0 < \( b \) < 1.
For the series \( \sum_{n=1}^{\infty} n^{\ln b} \), convergence depends on the sign of \( \ln b \):
- When \( \ln b < 0 \), \( n^{\ln b} \) eventually reaches zero as \( n \) goes to infinity, thus the series converges.
- Conversely, if \( \ln b \geq 0 \), \( n^{\ln b} \) fails to shrink to zero, causing divergence.
Therefore, the primary convergence condition is established by requiring \( b < 1 \) for convergence, ensuring \( \ln b \) is negative. This delicate balance of conditions allows us to deduce that the values of \( b \) which result in convergence are those where 0 < \( b \) < 1.
Logarithmic Functions
Understanding logarithmic functions is key in analyzing the series and finding its convergence conditions. Logarithms, often represented by \( \ln x \) for the natural logarithm, provide a mathematical tool for converting multiplicative scenarios into additive ones, which simplifies complex expressions greatly.
In our series, \( b^{\ln n} \) employs a logarithmic transformation to become \( n^{\ln b} \). This transformation is critical in testing the limit of this expression and understanding the convergence behavior. The key properties of logarithms here are:
For the convergence analysis, we observe how \( \ln b < 0 \) leads to a decreasing sequence, especially as \( n \) becomes large. This negative logarithm effectively reverses the expected growth, ensuring that the terms diminish in value and permitting series convergence. Thus, logarithmic functions serve as a bridge in converting exponential expressions into ones we can algebraically manipulate for deeper insights.
In our series, \( b^{\ln n} \) employs a logarithmic transformation to become \( n^{\ln b} \). This transformation is critical in testing the limit of this expression and understanding the convergence behavior. The key properties of logarithms here are:
- \( \ln(ab) = \ln a + \ln b \)
- \( \ln(a^b) = b\ln a \)
For the convergence analysis, we observe how \( \ln b < 0 \) leads to a decreasing sequence, especially as \( n \) becomes large. This negative logarithm effectively reverses the expected growth, ensuring that the terms diminish in value and permitting series convergence. Thus, logarithmic functions serve as a bridge in converting exponential expressions into ones we can algebraically manipulate for deeper insights.