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Magazine Chapter 9: Problem 8
Use comparison with \(\int_{1}^{\infty} x^{-3} d x\) to show that \(\sum_{n=2}^{\infty} 1 / n^{3}\) converges to a number less than or equal to \(1 / 2.\)
Short Answer
Expert verified
The series converges to a number \( \leq \frac{1}{2} \).
Step by step solution
01
Identify the given series
We are given the series \( \sum_{n=2}^{\infty} \frac{1}{n^{3}} \). We need to show that it converges, and compare it to an integral to find an upper bound.
02
Write the comparison integral
Consider the integral \( \int_{1}^{\infty} \frac{1}{x^{3}} \, dx \) as a comparison. This will help us determine the convergence and an upper bound for the series.
03
Evaluate the comparison integral
Calculate the integral:\[ \int_{1}^{\infty} \frac{1}{x^{3}} \, dx = \left[ -\frac{1}{2x^2} \right]_{1}^{\infty} = \lim_{b \to \infty} \left( 0 + \frac{1}{2} \right) = \frac{1}{2}. \]So, the integral converges to \( \frac{1}{2} \).
04
Apply the integral test
Since \( f(x) = \frac{1}{x^{3}} \) is a positive, continuous, and decreasing function for \( x \geq 1 \), the integral test states that the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3}} \) converges if \( \int_{1}^{\infty} \frac{1}{x^{3}} \, dx \) converges.
05
Show series part and comparison
The series \( \sum_{n=2}^{\infty} \frac{1}{n^{3}} \) is a part of the series \( \sum_{n=1}^{\infty} \frac{1}{n^{3}} \), which converges by the integral test. Therefore, \( \sum_{n=2}^{\infty} \frac{1}{n^{3}} \leq \int_{1}^{\infty} \frac{1}{x^{3}} \, dx = \frac{1}{2} \).
06
Conclude the upper bound
Since \( \sum_{n=1}^{\infty} \frac{1}{n^{3}} \) converges and \( \int_{1}^{\infty} \frac{1}{x^{3}} \, dx \) provides an upper bound of \( \frac{1}{2} \), it follows that the series \( \sum_{n=2}^{\infty} \frac{1}{n^{3}} \) also converges to a value less than or equal to \( \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Test for Series Convergence
The integral test is a fundamental tool for determining the convergence of infinite series. It works by comparing a series with an improper integral. To apply the integral test, you must first ensure that the function derived from the series terms is:
If these conditions hold true for a function \(f(x)\) and you have a series \(\sum_{n=1}^{\infty} a_n\) such that \(a_n = f(n)\), then there is a strong connection:
This means the behavior of the improper integral directly impacts the convergence of the series, making it a powerful comparison tool.
- Positive
- Continuous
- Decreasing
If these conditions hold true for a function \(f(x)\) and you have a series \(\sum_{n=1}^{\infty} a_n\) such that \(a_n = f(n)\), then there is a strong connection:
- If the integral \(\int_{1}^{\infty} f(x) \, dx\) converges, then the series \(\sum_{n=1}^{\infty} a_n\) also converges.
- If the integral diverges, so does the series.
This means the behavior of the improper integral directly impacts the convergence of the series, making it a powerful comparison tool.
Comparison Integral for Series Analysis
The comparison integral helps you analyze and find bounds for infinite series by evaluating a related integral. It is particularly useful when it is challenging to evaluate a series directly.
For instance, consider the series \(\sum_{n=2}^{\infty} \frac{1}{n^3}\) and the integral \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\). Here, the comparison integral illustrates two crucial aspects:
For instance, consider the series \(\sum_{n=2}^{\infty} \frac{1}{n^3}\) and the integral \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\). Here, the comparison integral illustrates two crucial aspects:
- Convergence: Since the integral \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\) converges, it implies the convergence of the related series.
- Upper Bound: The value of the integral, \(\frac{1}{2}\), provides an upper bound for the series, ensuring the series cannot sum to more than \(\frac{1}{2}\).
Working with Definite Integrals in Series Convergence
A definite integral calculates the area under a curve from one point to another, typically from \(a\) to \(b\). In the context of convergence analysis, it helps compare a series with the behavior of a continuous function over an interval. Here's how you use definite integrals effectively:
In our example, solving the integral \(-\frac{1}{2x^2} \bigg|_{1}^{b}\) and taking the limit as \(b \to \infty\) leads to \(\frac{1}{2}\). This outcome shows that even once the definite integral is solved, its finite result (\(\frac{1}{2}\)) affirms both the series' convergence and gives insight into its total sum being less than this bound.
- Defining the Function: Start by matching the series to an analogous function \(f(x)\), ensuring that this function is appropriate for the convergence conditions mentioned.
- Solving the Integral: Calculate the definite integral \(\int_{1}^{\infty} f(x) \, dx\) to evaluate convergence. For example, evaluating \(\int_{1}^{\infty} \frac{1}{x^3} \, dx\) informs us about the related series.
In our example, solving the integral \(-\frac{1}{2x^2} \bigg|_{1}^{b}\) and taking the limit as \(b \to \infty\) leads to \(\frac{1}{2}\). This outcome shows that even once the definite integral is solved, its finite result (\(\frac{1}{2}\)) affirms both the series' convergence and gives insight into its total sum being less than this bound.
