Question

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{\pi^{\ln n}}{(\ln n)^{n}} .\) Show that \(\frac{a_{2 n}}{a_{n}} \rightarrow 0\) as \(n \rightarrow \infty\).

Short answer

The limit of the ratio is zero, so \( \sum a_n \) converges.

Step-by-step solution

Setup the problem

We are given the sequence \( a_n = \frac{\pi^{\ln n}}{(\ln n)^{n}} \). We aim to evaluate \( \lim_{n \to \infty} \frac{a_{2n}}{a_n} \) and show that it approaches zero.

Substitute for \( a_{2n} \) and \( a_n \)

Substitute \( a_{2n} \) and \( a_n \) into the ratio. We have: \( a_{2n} = \frac{\pi^{\ln(2n)}}{(\ln(2n))^{2n}} \) and \( a_n = \frac{\pi^{\ln n}}{(\ln n)^{n}} \). Therefore, \( \frac{a_{2n}}{a_n} = \frac{\pi^{\ln(2n)}}{(\ln(2n))^{2n}} \cdot \frac{(\ln n)^{n}}{\pi^{\ln n}} \).

Simplify the ratio

Start simplifying the expression: \( \frac{a_{2n}}{a_n} = \frac{\pi^{\ln(2n)-\ln n} \cdot (\ln n)^{n}}{(\ln(2n))^{2n}} = \frac{\pi^{\ln 2 + \ln n - \ln n} \cdot (\ln n)^{n}}{(\ln(2n))^{2n}} = \frac{\pi^{\ln 2} \cdot (\ln n)^{n}}{(\ln(2n))^{2n}} \).

Further simplification

Express \( \ln(2n) \) as \( \ln 2 + \ln n \) then further simplify: \( \frac{a_{2n}}{a_n} = \frac{\pi^{\ln 2} \cdot (\ln n)^{n}}{(\ln 2 + \ln n)^{2n}} \).

Focus on applying limits

Consider the behavior of \( (\ln 2 + \ln n)^{2n} \) as \( n \to \infty \): since \( \ln 2 + \ln n \approx \ln n \) for large \( n \), this simplifies the dominant term to being \( (2\ln n)^n = (\ln n)^{2n} \).

Apply limit

Evaluate the limit: \[\lim_{n \to \infty} \frac{\pi^{\ln 2} \cdot (\ln n)^{n}}{(\ln(2n))^{2n}} = \lim_{n \to \infty} \frac{\pi^{\ln 2}}{(\ln n)^n} = \lim_{n \to \infty} \pi^{\ln 2} \cdot (\ln n)^{-n}. \]As \( n \to \infty \), \( (\ln n)^{-n} \) tends to zero, thus the entire expression approaches zero.

Key concepts

Generalized Ratio Test

The generalized ratio test is an extension of the fundamental ratio test. It is a valuable tool for examining the convergence or divergence of series where traditional methods might fall short. To use this test, you calculate the limit of the ratio of consecutive terms of a series as it approaches infinity. Specifically:

• If \( \lim_{n \to \infty} \frac{a_{2n}}{a_n} < \frac{1}{2} \), the series \( \sum a_n \) converges.
• If \( \lim_{n \to \infty} \frac{a_{2n+1}}{a_n} > \frac{1}{2} \), the series diverges.

This refinement helps in cases where the standard ratio test might not have enough power to resolve convergence, particularly in complex or oscillating sequences. In the given exercise, the test helps determine behavior by taking limits of complex expressions.

Advanced Calculus Problems

Advanced calculus problems often involve intricate sequences and challenging series. The depth of these problems lies in their complexity and the precision required to solve them.
Convergence of series is a recurring theme: understanding whether an infinite chain of numbers leads to a sum or not is crucial in many applications.The given problem with the series \( a_n = \frac{\pi^{\ln n}}{(\ln n)^n} \) presents a challenge because of its non-trivial components:

• The exponential term \( \pi^{\ln n} \) introduces non-linear growth.
• The divisor \( (\ln n)^n \) increases rapidly with \( n \).

Resolving the behavior of such a series through limits of product ratios pushes the boundary of understanding, requiring precision in setting up substitutions and simplifications.

Limits and Sequences

Limits and sequences are the foundation of calculus, providing a powerful method for analyzing the behavior of functions and series as they approach infinity.Sequence behavior is particularly vital in evaluating limits:

• The limit \( \lim_{n \to \infty} \frac{a_{2n}}{a_n} \) identifies how terms in the sequence behave relatively as \( n \) increases.
• In the given problem, simplification leads to understanding the dominant components of the sequence, revealing how terms headed towards zero indicates convergence.

Calculating these limits requires recognizing dominant behaviors in expressions, simplifying intricate parts, and sometimes using logarithmic properties for tractability.

Series Divergence

Series divergence occurs when the sum of the infinite terms does not settle at a finite value. This notion is crucial as not all series settle into a limit, some continue growing without bound.When applying the generalized ratio test, it's essential to check the conditions for divergence. In the exercise, a divergence condition would be modeled by:

• The limit \( \lim_{n \to \infty} \frac{a_{2n+1}}{a_n} \) appearing to be greater than \( \frac{1}{2} \). However, here it was established that the terms trend to zero, indicating no divergence was present.

Detecting divergence requires meticulously calculating the limits, ensuring no computational missteps, which fortifies your understanding of series behavior.