Problem 1 - Essential Calculus Early Transcendentals

Question

Draw a picture to show that $$\sum_{n=2}^{\infty} \frac{1}{n^{1.3}}<\int_{1}^{\infty} \frac{1}{x^{1.3}} d x$$ What can you conclude about the series?

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \) converges by integral comparison.

Step 1: Understand the Problem

We need to show that the sum of the series from \( n=2 \) to \( \infty \) of \( \frac{1}{n^{1.3}} \) is less than the integral from 1 to \( \infty \) of \( \frac{1}{x^{1.3}} \) with respect to \( x \). This will involve understanding both the discrete and continuous cases.

Step 2: Interpret the Series and Integral

Recognize that both the series \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \) and the integral \( \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \) represent sums of decreasing functions. The integral can often be used to approximate or bound the sum of a series, which is the strategy we will employ here.

Step 3: Visualize with a Graph

Draw the graph of \( f(x) = \frac{1}{x^{1.3}} \). Plot points for each integer \( n \), illustrating that each term \( \frac{1}{n^{1.3}} \) is less than the corresponding area under the curve from \( n \) to \( n+1 \). The series corresponds to all these discrete points, while the integral represents the continuous area under the curve.

Step 4: Setup the Integral Comparison

We will compare the sum \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \) to the integral \( \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \). Notice that the integral from 1 to \( \infty \) encompasses more area than just starting the series sum from 2, suggesting that the integral is greater.

Step 5: Calculate the Integral

Evaluate the improper integral \( \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \). The antiderivative of \( \frac{1}{x^{1.3}} \) is \( -\frac{1}{0.3} x^{-0.3} \), evaluated from 1 to \( \infty \). This becomes \( -\frac{1}{0.3}(x^{-0.3}) \bigg|_1^{\infty} = \frac{1}{0.3} \).

Step 6: Conclude Dominance

Comparing the results: since the series and the integral have the same function but different bounds, and because \( f(x) = \frac{1}{x^{1.3}} \) is a decreasing function, the integral will dominate the series. Hence, \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} < \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \).

Step 7: Convergence Conclusion

Since the integral \( \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \) converges, this establishes that the series \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \) also converges by comparison to the integral.

Key Concepts

Convergence of Series

When we talk about the **convergence of series**, we're describing whether a sequence of partial sums of the series approaches a definite value. For a series \( \sum_{n=1}^{\infty} a_n \), it converges if the limit of the sequence of its partial sums \( \{ S_N \} \), with \( S_N = a_1 + a_2 + \cdots + a_N \), approaches a finite number as \( N \) goes to infinity.

Convergence of series is important because it allows us to sum an infinite number of terms into a single finite value. Not all series converge—some may diverge, meaning their sums grow without bound or oscillate between different values.

For the series \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \), we are investigating whether adding these terms results in a finite sum or not. By comparing this series to a known convergent expression like an integral, we can determine its behavior.

Series and Sequences

**Series and sequences** are fundamental concepts in calculus and analysis. A sequence is an ordered list of numbers, defined explicitly or implicitly by a formula. For example, the sequence \( a_n = \frac{1}{n^{1.3}} \) produces terms like \( \frac{1}{2^{1.3}}, \frac{1}{3^{1.3}}, \ldots \).

A series, on the other hand, is the sum of a sequence. Consider the series formed by the sequence \( a_n \) from above: \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \). The series adds together every term of the sequence starting from \( n=2 \).

Understanding the behavior of a sequence provides insight into the behavior of its series. A sequence must approach zero for a series to even have the potential to converge, but this alone does not guarantee convergence. To establish convergence, we often use tests like the integral test.

Integral Test for Convergence

The **integral test for convergence** is a tool used to determine if a series converges. It leverages the connection between the sum of a series and an associated improper integral. Here's how it works:

• If \( a_n \) is positive, continuous, and decreasing for \( n \geq k \) (with \( k \) being a positive integer), then we can evaluate the improper integral \( \int_{k}^{\infty} f(x) \, dx \), where \( f(n) = a_n \).
• If the integral converges, then the series \( \sum_{n=k}^{\infty} a_n \) converges.
• Conversely, if the integral diverges, the series also diverges.

For example, with our exercise, we consider \( f(x) = \frac{1}{x^{1.3}} \). Since the integral \( \int_{1}^{\infty} \frac{1}{x^{1.3}} \, dx \) converges to a finite value, the integral test confirms that the series \( \sum_{n=2}^{\infty} \frac{1}{n^{1.3}} \) also converges.

The integral test is especially useful for functions and series with decreasing properties. It provides a direct method to evaluate convergence, making it a handy tool in many calculus problems.