Problem 39 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \operatorname{sech} n $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \operatorname{sech} n \) converges.

Step 1: Understand the Series

The series provided is \( \sum_{n=1}^{\infty} \operatorname{sech} n \), where \( \operatorname{sech} n = \frac{1}{\cosh n} = \frac{2}{e^n + e^{-n}} \). One should recognize that \( \operatorname{sech} n \) is a term involving the hyperbolic secant function.

Step 2: Analyze Term Behavior

Examine the behavior of \( \operatorname{sech} n \) as \( n \to \infty \). Since \( \cosh n = \frac{e^n + e^{-n}}{2} \), we have \( \operatorname{sech} n = \frac{2}{e^n + e^{-n}} \). As \( n \to \infty \), \( e^n \) dominates \( e^{-n} \), making \( \operatorname{sech} n \approx \frac{2}{e^n} \), which tends to zero as \( n \) increases.

Step 3: Apply the Comparison Test

Compare the series \( \sum_{n=1}^{\infty} \operatorname{sech} n \) to the series \( \sum_{n=1}^{\infty} \frac{2}{e^n} \). Notice that \( \operatorname{sech} n < \frac{2}{e^n} \). The series \( \sum \frac{2}{e^n} \) is a geometric series with ratio \( r = \frac{1}{e} < 1 \), which converges. By the Comparison Test, if a series with larger terms converges, then the original series also converges.

Step 4: Conclude Convergence

Since \( \sum \frac{2}{e^n} \) converges and \( \operatorname{sech} n < \frac{2}{e^n} \) for all \( n \), it follows that \( \sum_{n=1}^{\infty} \operatorname{sech} n \) also converges by the Comparison Test.

Key Concepts

Comparison Test

In mathematics, determining whether a given infinite series converges or diverges can be complex. One handy method is the comparison test. It is a tool used to compare a given series with a known series to draw conclusions about its behavior.
The principle behind the comparison test is simple:

• If a series \( \sum a_n \) converges and another series \( \sum b_n \) satisfies \( 0 \leq b_n \leq a_n \) for all \( n \), then \( \sum b_n \) also converges.
• Similarly, if \( \sum a_n \) diverges and \( b_n \geq a_n \), then \( \sum b_n \) diverges too.

This test is beneficial because it allows you to rely on the known convergence properties of one series to understand another, especially when the given series isn't straightforward to evaluate directly.
In our exercise involving the series \( \sum \operatorname{sech} n \), we used this test by comparing it to a geometric series. This approach is instrumental in establishing the convergence of the original series efficiently.

Geometric Series

Understanding geometric series is crucial in calculus, particularly when determining convergence. A geometric series is characterized by each term being a constant multiple, known as the common ratio \( r \), of the previous term.
Here's the standard form:

• \( \sum_{n=0}^{\infty} ar^n = a(1 + r + r^2 + r^3 + \ldots) \)

Such series exhibit fascinating properties:

• If \( |r| < 1 \), the series converges to \( \frac{a}{1-r} \).
• If \( |r| \geq 1 \), the series diverges.

In the example series \( \sum \frac{2}{e^n} \), we have a geometric series where \( a = 2 \) and \( r = \frac{1}{e} \), which is less than 1. Thus, it converges.
This property of geometric series is crucial for using them effectively as benchmarks in the comparison test, as demonstrated in our solution.

Hyperbolic Functions

Hyperbolic functions, such as \( \operatorname{sech} \), exhibit intriguing parallels to trigonometric functions, albeit with significant differences.
The hyperbolic secant function, \( \operatorname{sech} n \), is defined as:

• \( \operatorname{sech} n = \frac{1}{\cosh n} \)
• \( \cosh n = \frac{e^n + e^{-n}}{2} \)

As \( n \to \infty \), the term \( e^n \) grows much faster than \( e^{-n} \), making \( \cosh n \sim \frac{e^n}{2} \). Therefore, \( \operatorname{sech} n \approx \frac{2}{e^n} \), suggesting it involves rapidly decreasing terms.
The rapid decay in magnitude is why hyperbolic functions like \( \operatorname{sech} \) become essential in convergence analysis, allowing us to compare them with well-known series to determine their convergence properties effectively.