Problem 50 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{\tanh n}{n^{2}}\end{equation}

Step-by-Step Solution

Answer

The series converges.

Step 1: Understand the Series

We need to identify whether the series \( \sum_{n=1}^{\infty} \frac{\tanh n}{n^{2}} \) converges or diverges. The terms of this series are \( \frac{\tanh n}{n^{2}} \), where \( \tanh n \) is the hyperbolic tangent of \( n \).

Step 2: Approximate \( \tanh n \) for Large \( n \)

The function \( \tanh n = \frac{e^n - e^{-n}}{e^n + e^{-n}} \) approaches 1 as \( n \to \infty \). Thus, for large values of \( n \), \( \tanh n \approx 1 \).

Step 3: Analyze the Series' Behavior

Since \( \tanh n \approx 1 \) for large \( n \), the series behaves like \( \sum_{n=1}^{\infty} \frac{1}{n^2} \).

Step 4: Apply the p-Series Test

A p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \). In our case, \( p = 2 \), which means \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges. Thus the original series \( \sum_{n=1}^{\infty} \frac{\tanh n}{n^{2}} \) is expected to converge similarly.

Step 5: Conclude Using the Comparison Test

Since \( \frac{\tanh n}{n^{2}} \leq \frac{1}{n^{2}} \) for all \( n \), and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, by the Comparison Test, \( \sum_{n=1}^{\infty} \frac{\tanh n}{n^{2}} \) also converges.

Key Concepts

p-Series Test

The p-Series Test is a fundamental tool in determining the convergence of series that fit the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). This test states that such a series converges if the exponent \( p \) is greater than 1. Conversely, if \( p \leq 1 \), the series diverges.
For instance, the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) has \( p = 2 \). Since 2 is greater than 1, we can confidently say the series converges.
The power of this test lies in its simplicity and ability to quickly analyze series where each term diminishes based on a power of \( n \). You can apply the p-Series Test to recognize convergence or divergence even without knowing the precise behavior of more complex terms such as \( \tanh n \) in the exercise discussed.

Comparison Test

The Comparison Test is another handy tool for evaluating the convergence of series. It works by comparing a series in question to another one with known behavior, usually simpler and well-understood.
To use this test, you need two series: \( \sum a_n \) and \( \sum b_n \). If \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges. On the other hand, if \( b_n \leq a_n \) and \( \sum b_n \) diverges, \( \sum a_n \) diverges as well.
In the problem provided, the series \( \sum_{n=1}^{\infty} \frac{\tanh n}{n^2} \) can be compared to the simpler p-series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which converges. Since \( \frac{\tanh n}{n^2} \leq \frac{1}{n^2} \) for all \( n \), the Comparison Test confirms that the original series converges too.

Hyperbolic Functions

Hyperbolic functions, like \( \tanh x \), are analogous to the trigonometric functions but for hyperbolas instead of circles. They have unique properties that make them useful in calculus and series analysis.
The hyperbolic tangent function \( \tanh x \) is defined as \( \frac{e^x - e^{-x}}{e^x + e^{-x}} \). As \( x \) becomes very large, \( \tanh x \) approaches 1, which makes it behave predictably in terms of series convergence.
Understanding this behavior is crucial when analyzing series involving \( \tanh n \). For large \( n \), \( \tanh n \approx 1 \), thus simplifying the series \( \sum_{n=1}^{\infty} \frac{\tanh n}{n^2} \) to resemble \( \sum_{n=1}^{\infty} \frac{1}{n^2} \). This approximation helps in using the p-Series and Comparison Tests to determine convergence.