Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give conditions on \(p\) that determine the convergence or divergence of \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\left(1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\frac{1}{n^{p}}\right)\).

Step-by-Step Solution

Answer

The series converges for \(p > 1\) and diverges for \(p \leq 1\).

Step 1: Understand the Problem

We need to determine the conditions on the parameter \(p\) under which the given series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\left(1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\frac{1}{n^{p}}\right)\) converges or diverges.

Step 2: Simplify the Inner Sum

The term inside the series is an inner sum \(1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{n^p}\). This is a truncated version of the Harmonic series of order \(p\) and is asymptotically equivalent to \(\zeta(p) - \sum_{k=n+1}^{\infty} \frac{1}{k^p}\) as \(n \to \infty\), where \(\zeta(p)\) is the Riemann zeta function.

Step 3: Estimate the Inner Sum

Since the inner series converges to \( \sum_{k=1}^{n} \frac{1}{k^p} \sim \frac{n^{1-p}}{1-p} \) when \(p > 1\), we use the asymptotic approximation \(\sim \frac{n^{1-p}}{1-p}\) for large \(n\) and valid \(p\).

Step 4: Combine Inner and Outer Series

The outer series becomes \(\sum_{n=1}^{\infty} \frac{n^{1-p}}{n^p} \sim \sum_{n=1}^{\infty} n^{1-2p}\). The behavior of this series depends on the exponent \(1-2p\).

Step 5: Analyze Series Convergence

The series \(\sum_{n=1}^{\infty} n^{1-2p}\) converges if \(1-2p < -1\), which simplifies to \(2p > 2\) or \(p > 1\). Conversely, if \(p \leq 1\), then the series diverges.

Step 6: Conclude Conditions

The series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\left(1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\frac{1}{n^{p}}\right)\) converges if \(p > 1\) and diverges if \(p \leq 1\).

Key Concepts

Riemann zeta function

The Riemann zeta function, denoted as \( \zeta(p) \), is a fundamental tool in mathematical analysis, particularly in number theory and complex analysis. It is defined for complex numbers \( p \) with a real part greater than 1 and is defined as the infinite sum:

• \( \zeta(p) = \sum_{n=1}^{\infty} \frac{1}{n^p} \)

This function captures how sums of reciprocals of integer powers behave as we sum indefinitely.
It is crucial when analyzing the convergence of series, like the one in the problem statement, because it complements our understanding of how partial sums behave.For real numbers \( p > 1 \), the series defined by the zeta function converges due to the decreasing magnitude of the terms. When \( p = 1 \), this equates to the harmonic series which diverges, indicating the boundary at which convergence changes to divergence. The connection to the given problem lies in how the zeta function is used in an asymptotic approximation, helping us identify convergence conditions for complex expressions by simplifying into more known forms.

Harmonic series

The Harmonic series is a famous and simple divergent series, expressed as:

• \( \sum_{n=1}^{\infty} \frac{1}{n} \)

Despite its simple structure, it grows logarithmically towards infinity and fails to settle at a single sum, hence it diverges.
In higher powers, the series can be generalized as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) for \( p \leq 1 \), and these also demonstrate divergent behavior.
The harmonic series serves as a form of baseline or reference point for analyzing other series.In our scenario, the terms inside the sum resemble a generalization of the harmonic series for \( p \) in higher orders. When \( p > 1 \), convergence occurs, as mentioned in the Riemann zeta function explanation. This property goes hand in hand with convergence criteria analysis within the original exercise's bounds.

asymptotic approximation

Asymptotic approximation is a technique used in mathematics to describe the behavior of functions as a variable, typically related to size, approaches a certain limit, often infinity.In the exercise under discussion, we use asymptotic approximation to simplify the inner summation of the series:

• The sum \( 1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots + \frac{1}{n^p} \) is approximated by the asymptotic term \( \frac{n^{1-p}}{1-p} \) when \( p > 1 \).

Understanding this asymptotic behavior is crucial since it is part of determining when the original series will converge. By using asymptotic approximations, complex expressions involving infinite sums can be transformed into simpler forms.
These approximations highlight the leading terms that capture the core behavior of functions or sequences and assist in unveiling convergence conditions efficiently.