Problem 20 - Essential Calculus

Question

Use the Integral Test to determine the convergence or divergence of the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 / 3}} $$

Step-by-Step Solution

Answer

The infinite p-series \( \sum_{n=1}^{\infty} 1/n^{1 / 3}\) diverges.

Step 1: Identify the series type and function

Identify the function in the given series. The series is a p-series and the function is \( f(n)= 1/n^{1 / 3}\).

Step 2: Demonstrate criteria to apply Integral Test

Show that the function f(n)= 1/n^{1 / 3} is continuous, positive, and decreasing on [1, ∞). Since \(n^{1 / 3}\) is a cube root function, it is continuous and positive for \(n \geq 1\), and it's decreasing as the denominator increases: the larger the \(n\), the smaller the \(f(n)\). Therefore, the conditions to apply the Integral Test are met.

Step 3: Apply the Integral Test

Take the integral from 1 to ∞ of the function \(f(x)= 1/x^{1 / 3}\) to determine if it converges or diverges. The integral is \(\int_{1}^{\infty} x^{-1 / 3} dx\). Evaluate the integral to see if it converges or diverges. The antiderivative is \(3/2*x^{2 / 3}\), and evaluating from 1 to ∞ gives \(\infty\) - \(3/2\) which is \(\infty\). Therefore, the integral diverges.

Key Concepts

Understanding p-Series

A p-series is a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Analyze it by looking at the exponent \( p \). If \( p \) is greater than one, the series converges. Conversely, when \( p \leq 1 \), the series diverges. This specific exercise features \( p=1/3 \).Key characteristics of p-series include:

• The series has the general term \( \frac{1}{n^p} \).
• It is evaluated based on the value of \( p \).
• It can either converge or diverge based on \( p \).

Given the form \( \sum_{n=1}^{\infty} \frac{1}{n^{1/3}} \), the exponent \( p = 1/3 \) indicates that the series will diverge because \( p < 1 \). Applying the Integral Test confirms this observation, linking continuous functions with their corresponding series.

Convergence and Divergence Explained

Convergence and divergence of series are key concepts in evaluating infinite series. Convergence means the series sums to a finite number. Divergence implies it does not. For the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), understanding convergence or divergence depends on the value of \( p \).Reasons why a series converges or diverges:

• A series converges if the sum of its terms approaches a specific value.
• It diverges if the sum grows indefinitely.

The series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/3}} \), has \( p = 1/3 \), clearly diverges since \( p \leq 1 \). Applying tests like the Integral Test helps determine the behavior of such series by connecting their terms to integrals.

Improper Integrals in the Integral Test

Improper integrals are used to evaluate the convergence of infinite series. An improper integral is one where the limits of integration are infinite, or the integrand has a vertical asymptote.Steps to evaluate an improper integral:

• Express the integral \( \int_{a}^{b} f(x) \, dx \), where \( a \) or \( b \) can be infinity.
• Find the antiderivative of the function \( f(x) \).
• Calculate the limit as one or both bounds approach infinity.

When applying the Integral Test to the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1/3}} \), one considers the integral \( \int_{1}^{\infty} \frac{1}{x^{1/3}} \ dx \). Solving this integral helps determine the convergence of the series. Here, the antiderivative \( \frac{3}{2}x^{2/3} \) evaluated at the bounds reveals an infinite result, affirming the series's divergence.