Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(\sum a_{n}\) is a convergent series of positive terms then \(\sum \ln \left(1+a_{n}\right)\) converges.

Step-by-Step Solution

Answer

The series \(\sum \ln(1 + a_{n})\) converges by the Comparison Test.

Step 1: Understand the Given Problem

We have a series \(\sum a_{n}\) that is convergent with positive terms. We need to prove that the series \(\sum \ln(1 + a_{n})\) also converges.

Step 2: Use Series Convergence Criteria

For a series \(\sum a_{n}\) to converge, it is necessary for \(a_{n} \to 0\) as \(n \to \infty\). This holds true since \(a_{n}\) are positive terms and the series is convergent.

Step 3: Apply the Logarithmic Approximation

For small \(x\), \(\ln(1 + x) \approx x\). Since \(a_{n} \to 0\), for sufficiently large \(n\), \(\ln(1 + a_{n}) \approx a_{n}\).

Step 4: Determine Convergence Using the Comparison Test

Since \(\ln(1 + a_{n}) \approx a_{n}\) for large \(n\) and \(\sum a_{n}\) converges, \(\sum \ln(1 + a_{n})\) will also converge by the Comparison Test. This is because \(\ln(1 + a_{n})\) behaves similarly to \(a_{n}\) for large \(n\), and is certainly smaller than some constant multiple of \(a_{n}\), ensuring convergence.

Key Concepts

Comparison Test

The Comparison Test is a powerful tool in the analysis of series convergence. It allows us to compare a series of interest with a simpler series, whose convergence properties we already know. If a series \( \sum a_{n} \) converges and \( \ln(1 + a_{n}) \approx a_{n} \) for large \( n \), then the series \( \sum \ln(1 + a_{n}) \) can be compared to the convergent series \( \sum a_{n} \).
This similarity implies that \( \sum \ln(1 + a_{n}) \) inherits the convergence from \( \sum a_{n} \) because the logarithmic terms are sufficiently small and closely related to the terms \( a_n \).
This test is essential when dealing with series that are not easily summable outright, but can be shown to behave similarly to a known convergent series.

Logarithmic Approximation

A key aspect in the convergence proof is using a logarithmic approximation. This is particularly useful when considering terms in a series where the values of \( a_{n} \) become very small. For small values of \( x \), the natural logarithm \( \ln(1 + x) \) can be approximated as \( x \).
This approximation is based on the initial term in the Taylor series expansion of \( \ln(1 + x) \).

• For small \( x \), higher-order terms in the expansion become negligible.
• This means \( \ln(1 + a_{n}) \approx a_{n} \).

This simplification plays a vital role in proving that \( \sum \ln(1 + a_{n}) \) behaves much like \( \sum a_{n} \), ensuring that the convergence properties are essentially equivalent for the purposes of analysis.

Positive Terms in Series

When working with a series \( \sum a_{n} \) where each term \( a_{n} \) is positive, the terms' positivity is crucial in applying convergence tests. Positive terms imply that each successive partial sum is larger, or at least not smaller, than the preceding one, providing a solid basis for certain analytical techniques like the Comparison Test.
A series composed of positive terms is non-decreasing in its partial sums, ensuring that oscillatory behavior does not obscure convergence properties.

• The behavior of positive terms leads to two potential outcomes: either the series reaches a finite limit, indicating convergence, or it diverges.
• For a convergent series of positive terms \( \sum a_{n} \), it's indeed necessary that \( a_{n} \to 0 \) as \( n \to \infty \), ensuring that additional terms add progressively smaller values.

This foundational understanding aids in linking the behavior of \( \sum a_{n} \) to that of \( \sum \ln(1 + a_{n}) \), making convergence proofs more straightforward.