Problem 46 - Essential Calculus Early Transcendentals

Question

Find all positive values of \(b\) for which the series \(\sum_{n=1}^{\infty} b^{\ln n}\) converges.

Step-by-Step Solution

Answer

The series converges for \(0 < b < \frac{1}{e}\).

Step 1: Understand the Series

The series in question is \(\sum_{n=1}^{\infty} b^{\ln n}\). The term \(b^{\ln n}\) can be rewritten using the property of exponents: \(b^{\ln n} = e^{(\ln b)(\ln n)} = n^{\ln b}\). Thus, the series is \(\sum_{n=1}^{\infty} n^{\ln b}\).

Step 2: Identify Convergence Criteria

We need to determine the condition for convergence of the series \(\sum_{n=1}^{\infty} n^{\ln b}\). A series of the form \(\sum_{n=1}^{\infty} n^{p}\) converges if \(p < -1\). This implies that for convergence, \(\ln b\) must satisfy \(\ln b < -1\).

Step 3: Solve the Inequality

To find the allowable values of \(b\), solve the inequality \(\ln b < -1\). This implies:\[ b < e^{-1} \]. Since we are looking for positive values of \(b\), this gives us the range \(0 < b < \frac{1}{e}\).

Step 4: Finalize the Solution

Given the inequality from the solution above, the series \(\sum_{n=1}^{\infty} b^{\ln n}\) converges for positive values of \(b\) such that \(0 < b < \frac{1}{e}\).

Key Concepts

Understanding Exponent Properties

The original series involves terms like \(b^{\ln n}\) which can initially seem puzzling. To handle this, we need to tap into the properties of exponents laid out in mathematics.One useful property is that when you have an expression like \(b^{\ln n}\), it can be rewritten using the natural exponent, \(e\):

• First, recognize that \(b^{\ln n} = e^{\ln(b^{\ln n})}\)
• Apply the Logarithmic Identity: \(\ln(b^{\ln n}) = (\ln n)(\ln b)\)
• Thus, \(b^{\ln n} = e^{(\ln b)(\ln n)}\)

With these properties, the term \(b^{\ln n}\) can be simplified further to another familiar form: \(n^{\ln b}\).This transformation is an essential part of understanding the series setup and helps when analyzing convergence. Knowing how to manipulate exponents is crucial for working with complex expressions in calculus and series.

Solving Inequalities for Convergence

The series \(\sum_{n=1}^{\infty} n^{\ln b}\) must converge for certain values of \(b\). To solve for these values, we must delve into inequalities. Convergence of a series \(\sum_{n=1}^{\infty} n^p\) depends on the value of \(p\). Specifically:

• The series converges if \(p < -1\).

Now that we have expressed the terms of our series as \(n^{\ln b}\), we equate \(\ln b\) with \(p\). Thus, the inequality that determines convergence becomes:\[\ln b < -1\] Solving this inequality is a matter of understanding the behavior of logarithms:

• Exponentiating both sides gives us: \(b < e^{-1}\).
• Since we are dealing with positive values for \(b\), this implies \(0 < b < \frac{1}{e}\).

This interval ensures the series converges. Inequality solving is a powerful tool in calculus for determining such crucial conditions.

Determining Positive Values

Once we've chosen the inequality \(\ln b < -1\), solving it tells us something vital: the set of possible positive values for \(b\).Identify the Range
Start with understanding \(\ln b < -1\). When you exponentiate to eliminate the logarithm, you derive the condition:

• \(b < e^{-1}\)

However, we're specifically interested in positive numbers.

• This means considering \(b > 0\) alongside \(b < e^{-1}\).

Hence, the possible positive range becomes:\(0 < b < \frac{1}{e}\).This range does more than ensure convergence; it sets boundaries for the behavior of the series itself.This determination provides a clear answer on the conditions under which our original series converges: it only converges when \(b\) lies within this specified range.