Problem 45 - Thomas Calculus

Question

Which of the series in Exercises 13 46 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 converges by the Comparison Test.

Step 1: Understand the Series

The series given is \( \sum_{n=1}^{\infty} \operatorname{sech} n \). The term \( \operatorname{sech} n \) stands for the hyperbolic secant function, which is defined as \( \operatorname{sech} n = \frac{1}{\cosh n} \). Since \( \cosh n > 1 \) for all \( n \geq 1 \), it follows that \( \operatorname{sech} n < 1 \).

Step 2: Apply the Comparison Test

To determine the convergence or divergence of the series, we can use the Comparison Test. We know that for large values of \( n \), \( \cosh n \approx \frac{e^n}{2} \), so \( \operatorname{sech} n \approx \frac{2}{e^n} \). This closely resembles a geometric series.

Step 3: Analyze a Related Series

Consider the series \( \sum_{n=1}^{\infty} \frac{2}{e^n} \). This is a geometric series with a ratio \( r = \frac{1}{e} \). Since \( r < 1 \), this geometric series converges.

Step 4: Conclude based on the Comparison Test

Since \( \operatorname{sech} n \approx \frac{2}{e^n} \) for large \( n \) and the series \( \sum \frac{2}{e^n} \) converges, we can conclude using the Comparison Test that the series \( \sum \operatorname{sech} n \) also converges.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, much like trigonometric functions, play a vital role in mathematics, encompassing areas such as calculus and complex analysis. They are analogs to the familiar sine, cosine, and tangent but for hyperbolas rather than circles. Among these functions, the hyperbolic secant, denoted as \( \operatorname{sech}\), is particularly important. It is defined as:

• \( \operatorname{sech} x = \frac{1}{\cosh x} \)
• Where \( \cosh x = \frac{e^x + e^{-x}}{2} \) is the hyperbolic cosine function

What makes hyperbolic functions unique is their connection to exponential functions, allowing them to model growth and decay processes efficiently.
Specifically, the hyperbolic secant function \( \operatorname{sech} x \) is always less than one for real \( x \), since \( \cosh x \geq 1 \).
In the realm of series convergence, such characteristics of hyperbolic functions help determine whether a series involving hyperbolic terms will converge or diverge.

Comparison Test

The Comparison Test is a powerful tool in determining the convergence or divergence of infinite series. It leverages the properties of a well-understood series to deduce the behavior of a more complicated one.
Here’s how it works:

• Suppose you have two series \( \sum a_n \) and \( \sum b_n \).
• If \( 0 \leq a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \geq 0 \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges.

Applying this to our problem, we approximate \( \operatorname{sech} n \) by \( \frac{2}{e^n} \) for large \( n \).
Since \( \sum \frac{2}{e^n} \) is a geometric series that converges (because its common ratio \( r = \frac{1}{e} \) is less than one),
the original series \( \sum \operatorname{sech} n \) also converges by comparison.

Geometric Series

A geometric series is one of the simplest and most fundamental types of series in mathematics. It's characterized by each term being a constant multiple, known as the common ratio, of the previous term. Geometric series take the form:

• \( \sum_{n=0}^{\infty} ar^n \)
• Where \( a \) is the first term and \( r \) is the common ratio.

The convergence of a geometric series depends solely on that common ratio:

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

In the context of our exercise, the series \( \sum \frac{2}{e^n} \) is a geometric series with first term \( a = 2 \) and ratio \( r = \frac{1}{e} \), which is less than one.
This means the series converges, helping us infer via the Comparison Test
that the series \( \sum \operatorname{sech} n \) converges as well.