Problem 47 - 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{\tan ^{-1} n}{n^{1.1}}\end{equation}

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} \) converges.

Step 1: Identify the Type of Series

We aim to determine the convergence of the series \( \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} \). This series involves the arctan function over a polynomial of \( n \), which suggests it might be comparable to a p-series once analyzed.

Step 2: Examine the Series' Behavior

Since \( \tan^{-1}(n) \) approaches \( \frac{\pi}{2} \) as \( n \to \infty \), the terms \( \frac{\tan^{-1}(n)}{n^{1.1}} \) behave similarly to \( \frac{\frac{\pi}{2}}{n^{1.1}} \) for large \( n \). This comparison suggests the series shares similar properties with a p-series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \).

Step 3: Apply the Comparison Test

Consider the function \( \lim_{n \to \infty} \frac{\tan^{-1}(n)}{n^{1.1}} \times \frac{n^{1.1}}{1} = \lim_{n \to \infty} \frac{\tan^{-1}(n)}{1} \). Since \( \tan^{-1}(n) \) is bounded and approaches \( \frac{\pi}{2} \), compare \( \frac{\tan^{-1}(n)}{n^{1.1}} \) to \( \frac{1}{n^{1.1}} \), a convergent p-series (because \( p = 1.1 > 1 \)). The comparison test shows that if the p-series converges, then our original series converges as well.

Step 4: Conclude the Series' Behavior

Since \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) converges, and \( \frac{\tan^{-1}(n)}{n^{1.1}} \leq \frac{C}{n^{1.1}} \) for some constant \( C \), the original series \( \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} \) converges by the comparison test.

Key Concepts

Comparison Test

The Comparison Test is a handy tool when you want to determine if a series converges or diverges. It works by comparing the series of interest with another series whose behavior (convergence or divergence) is already known.
For example, if you suspect your series is similar to a known convergent series, you can use the Comparison Test to prove convergence.

• Find a known series to compare with.
• Establish that the terms of your unknown series are less than or equal to those of the known series.
• If the known series converges and your series is smaller, then your series converges too.
• Conversely, if you establish terms are larger and the known series diverges, then your series diverges.

In our problem, we compare the original series with a p-series. Both share a similar structure, which makes the Comparison Test easy to apply.

p-Series

A p-Series is a type of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The behavior of p-Series is predictable based on the value of \( p \). Here's the essential information:

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In the original exercise, our series \( \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} \) was compared to a p-Series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \). Since \( p = 1.1 \) (where \( 1.1 > 1 \)), the p-Series converges. By the Comparison Test, the original series must also converge. Understanding the behavior of p-Series is crucial for applying this method effectively.

Arctan Function

The Arctan Function, denoted as \( \tan^{-1}(x) \), is the inverse of the tangent function. Understanding its behavior is important when analyzing series that include this function.

• As \( x \to \infty \), \( \tan^{-1}(x) \to \frac{\pi}{2} \). This is an asymptotic behavior known as a horizontal asymptote.
• The function is bounded and increases slowly as \( x \) grows.

Given this behavior, in our series \( \sum_{n=1}^{\infty} \frac{\tan^{-1}(n)}{n^{1.1}} \), as \( n \) becomes very large, \( \tan^{-1}(n) \) can be approximated by \( \frac{\pi}{2} \). This similarity allows us to replace \( \tan^{-1}(n) \) with a constant when comparing with a p-Series. Such insights simplify the analysis and help us apply convergence tests effectively.