Problem 19 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

The p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{3}}\) converges.

Step 1: Set up the Integral

According to the Integral Test, we compare the series to the integral \( \int_{1}^{\infty}\frac{1}{x^{3}}dx\)

Step 2: Calculate the Integral

We can calculate this integral as follows: \( \int_{1}^{\infty}\frac{1}{x^{3}}dx = [-\frac{1}{2x^{2}}]|_{1}^{\infty} \).

Step 3: Evaluate the Integral

First, plug in \( \infty \) and then subtract the result when \( 1 \) is plugged in: \( \lim_{t\to \infty}[-\frac{1}{2t^{2}}] + \frac{1}{2} \). Because as \( t \) approaches \( \infty \), \( \frac{1}{2t^{2}} \) approaches \( 0 \), this evaluates to \( 0 + \frac{1}{2} \).

Step 4: Determine Convergence or Divergence

Since the integral is finite, it converges. Therefore, according to the Integral Test, the series also converges.

Key Concepts

p-series

A p-series is a type of infinite series defined as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive real number. This series is particularly interesting because its behavior—whether it converges or diverges—depends heavily on the value of \( p \).
When \( p > 1 \), the p-series converges. Conversely, when \( p \leq 1 \), the p-series diverges. The reason for this distinction will become clearer when we discuss the Integral Test, which allows us to determine the convergence or divergence of this series based on the behavior of corresponding integrals.

convergence

Convergence, in the context of series, means that the series approaches a specific value as more and more terms are added. Essentially, no matter how many terms you add, the sum does not get infinitely large; it stabilizes at a finite number.

• The convergence of a series depends on the terms diminishing rapidly enough as the series progresses.
• For the p-series \( \sum_{n=1}^{\infty}\frac{1}{n^3} \), we observe convergence because each term \( \frac{1}{n^3} \) decreases quickly enough for \( p=3 \) (where \( p > 1 \)).
• This rapid reduction ensures the sum of the series remains finite, confirming convergence.

divergence

Divergence in a series means that as you add more terms, the series grows without bounds. It doesn't settle towards any specific value, and continues to increase indefinitely.

• Divergence often occurs when the terms of a series decrease at a slower rate, preventing the sum from reaching a stable, finite number.
• An example would be a p-series where \( p \leq 1 \). For instance, \( \sum_{n=1}^{\infty}\frac{1}{n} \) is a divergent series because the terms \( \frac{1}{n} \) do not diminish rapidly enough for the series to converge.
• Under the Integral Test, if the integral is infinite, the p-series is considered divergent.

Improper Integrals

An improper integral is an integral where either the interval of integration is infinite, or the function being integrated has an infinite discontinuity. These integrals are essential in analyzing the convergence of series through the Integral Test.

• In order to determine the convergence of the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \), the corresponding integral \( \int_{1}^{\infty}\frac{1}{x^3}dx \) needs to be evaluated.
• This integral is improper because the upper limit of integration is infinite.
• By evaluating this integral, we calculate whether it results in a finite or infinite value. If finite, the series converges; if infinite, it diverges.
• For the given series \( \sum_{n=1}^{\infty}\frac{1}{n^3} \), the integral evaluates to a finite value indicating convergence, which confirms the behavior of the series.