Problem 64 - Thomas Calculus

Question

(Continuation of Exercise \(63 . )\) Show that \(\sum_{n=2}^{\infty}\left((\ln n)^{q} / n^{p}\right)\) diverges for \(-\infty < q < \infty\) and \(0 < p < 1 .\) (Hint: Limit Comparison with an appropriate p-series.)

Step-by-Step Solution

Answer

The series diverges for \(-\infty < q < \infty\) and \(0 < p < 1\).

Step 1: Understand the Series

We are given the series \( \sum_{n=2}^{\infty}\frac{(\ln n)^q}{n^p} \). We need to determine whether this series diverges for \(-\infty < q < \infty\) and \(0 < p < 1\). To determine divergence, we will use the Limit Comparison Test with a p-series.

Step 2: Choose a Comparison Series

We chose the series \( \sum_{n=2}^{\infty} \frac{1}{n^p} \) to compare against our original series, because it resembles a p-series where \( p < 1 \), which is known to diverge.

Step 3: Apply the Limit Comparison Test

Calculate the limit \( \lim_{n \to \infty} \frac{(\ln n)^q / n^p}{1/n^p} = \lim_{n \to \infty} (\ln n)^q \).

Step 4: Analyze the Limit

The behavior of \( \lim_{n \to \infty} (\ln n)^q \) varies based on the value of \( q \) but importantly, for any real \( q \), the function \((\ln n)^q\) increases unbounded as \(n\) approaches infinity.

Step 5: Determine Divergence

Since \((\ln n)^q\) grows without bound, the calculated limit in Step 3 is infinite (or grows without bounds), thus the original series and \( \sum_{n=2}^{\infty} \frac{1}{n^p} \) both diverge by the Limit Comparison Test.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a tool used to determine the convergence or divergence of an infinite series. It works by comparing two series in such a way that if one series behaves in a known manner (convergent or divergent), the compared series will follow the same behavior.

Here's how it works:

• We look at two series: \( \sum a_n \) and \( \sum b_n \) , where both consist of positive terms.
• We compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If this limit is a positive finite value, then both series will either converge or diverge together.

In the exercise, the Limit Comparison Test is applied by choosing the comparison series \( \sum_{n=2}^{\infty} \frac{1}{n^p} \), which is a p-series known to diverge when \(0 < p < 1\). The test involves comparing our given series \( \sum_{n=2}^{\infty} \frac{(\ln n)^q}{n^p} \) against this known p-series. By examining the limit, we find it unbounded, aligning the series behaviors so that both diverge.

p-series

A p-series is an essential type of series in calculus defined as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The parameter \( p \) plays a critical role in the behavior of the series:

• When \( p > 1\), the series converges.
  
• When \( p \leq 1\), the series diverges.

This rule provides a quick reference to determine whether a p-series converges or diverges without complex calculations.

In the given exercise, the series \( \sum_{n=2}^{\infty} \frac{1}{n^p} \) is used as a comparison series, as it diverges when \(0 < p < 1\). This contrasts with the behavior of a convergent p-series, facilitating the use of the Limit Comparison Test to ascertain the divergence of the original series \( \sum_{n=2}^{\infty}\frac{(\ln n)^q}{n^p} \). By recognizing the diverging nature of the comparison series, it's easier to conclude that the original series also diverges.

Logarithmic Functions

Logarithmic functions, particularly \(\ln n\), are fundamental in mathematics for transforming multiplication into addition, among other uses. They have unique characteristics that make them useful in various types of series analysis:

• When \(q\) is positive, \((\ln n)^q\) powers up the logarithmic function, growing as \(n\) increases.
• If \(q\) is zero, the expression simplifies to 1, which does not affect the series significantly.
• When \(q\) is negative, it results in the reciprocal of the powered logarithmic function, still growing because \(\ln n\) increases as \(n\) grows.

In the exercise, the power of the logarithmic component \((\ln n)^q\) dictates the behavior as \(n\) approaches infinity.

Regardless of the value of \(q\) (positive, negative, or zero), \( (\ln n)^q \) exhibits growth without bounds. This unbounded nature contributes to the divergent behavior of the series \( \sum_{n=2}^{\infty} \frac{(\ln n)^q}{n^p} \) when analyzed with the Limit Comparison Test.