Problem 32 - Single Variable Calculus

Question

Determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\)

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{\tan^{-1}n}{n^3+1}\) converges by the Comparison Test with the convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), as we proved that \(\frac{\tan^{-1}n}{n^3+1} \leq \frac{1}{n^2}\) for \(n \geq 1\).

Step 1: Choose an appropriate series to compare with

Let's start by looking at the original series: \[\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\] Since the function \(\tan^{-1} n\) grows very slowly, we can compare it with a slower-growing function (e.g., a p-series with \(n^3\) in the denominator). So, let's consider the series: \[\sum_{n=1}^{\infty} \frac{1}{n^2}\]

Step 2: Apply the Comparison Test

Now, apply the Comparison Test to see if the original series converges. We want to show that if \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges, then \(\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\) converges too. We will prove that \(\frac{\tan ^{-1} n}{n^{3}+1} \leq \frac{1}{n^2}\) for any \(n \geq 1\). First, we know that \(\tan^{-1} n \leq n\) for all \(n \geq 1\), since the function \(\tan^{-1} n\) grows slower than the linear function of n. Using this inequality, we get: \[\frac{\tan ^{-1} n}{n^{3}+1} \leq \frac{n}{n^{3}+1}\] Now, let's show that \(\frac{n}{n^{3}+1} \leq \frac{1}{n^2}\): \[\frac{n}{n^{3}+1} \leq \frac{1}{n^2}\] \[n^3 \leq n^{3}+1\] This inequality is true for \(n \geq 1\).

Step 3: Conclude the results

Since \(\frac{\tan ^{-1} n}{n^{3}+1} \leq \frac{1}{n^2}\) for \(n \geq 1\), and we know that the series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges (it is a p-series with p = 2 > 1), we can conclude, by the Comparison Test, that the original series also converges: \[\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\] converges.

Key Concepts

Comparison Test

The Comparison Test is a fundamental tool for determining the convergence or divergence of an infinite series. This test helps us analyze a given series by comparing it to a second series whose convergence properties are already known. The central idea is:

• If a known convergent series is greater than or equal (term by term) to the series in question, the test assures us that the unknown series also converges.
• Conversely, if a divergent series is less than or equal (term by term) to the series in question, the unknown series must also diverge.

In this exercise, we compared the series \(\sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{3}+1}\) to the known convergent series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\). Given that \(\frac{\tan^{-1} n}{n^3+1}\leq\frac{1}{n^2}\), and the latter converges, it follows by the Comparison Test that the original series converges as well.
Understanding when and how to apply this test effectively can simplify solving convergence questions a lot.

P-Series

P-series are a specific type of infinite series expressed in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where \(p\) is a constant. They are particularly significant because their convergence properties are well established and straightforward:

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

The exercise leverages the convergence of the \(p\)-series with \(p = 2\) or \(\sum_{n=1}^{\infty} \frac{1}{n^2}\). This series is known to converge because \(2 > 1\). Understanding the properties of \(p\)-series becomes vital when you use the Comparison Test because it provides a benchmark for determining the behavior of more complex series.

Trigonometric Functions

Trigonometric functions include familiar functions like sine, cosine, and tangent. In this problem, we specifically deal with the inverse tangent function, denoted as \(\tan^{-1} n\) or arctan. It's important to recognize the properties of \(\tan^{-1} n\):

• It is a monotonically increasing function, growing slower than linear growth \(n\).
• As \(n\) increases, \(\tan^{-1} n\) approaches \(\frac{\pi}{2}\), making it bounded.

Understanding these properties allowed us to make the essential comparison \(\tan^{-1} n \leq n\), crucial for showing that \(\frac{\tan^{-1} n}{n^3+1}\) is less than or equal to \(\frac{1}{n^2}\). Recognizing the growth pattern and limits of trigonometric functions simplifies the process of tackling convergence in such series.

Infinite Series

Infinite series consist of an infinite sum of terms written in the form \(\sum_{n=1}^{\infty} a_n\). Their analysis is crucial in many mathematical and scientific applications when we want to know their convergence or divergence. Here's what you need to know:

• When a series converges, the sum approaches a finite limit as \(n\) increases indefinitely.
• If a series diverges, the sum either increases without bound or oscillates without settling.
• Understanding series convergence helps in approximation, calculations and solving real-world problems.

This exercise provides an example with the series \(\sum_{n=1}^{\infty} \frac{\tan^{-1} n}{n^3+1}\), where, through appropriate methods like the Comparison Test, convergence is determined. Mastery of infinite series concepts is fundamental for anyone diving deeper into the study of calculus and analysis.