Problem 47 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1.1}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1.1}} \). This is an infinite series where the general term is \( a_n = \frac{\tan^{-1} n}{n^{1.1}} \). The series appears to be a non-standard one, and we need to determine if it converges or diverges.

Step 2: Use Comparison Test

The function \( \tan^{-1} n \) is known to be bounded, meaning for all integers \( n \geq 1 \), \( 0 \leq \tan^{-1} n \leq \frac{\pi}{2} \). Thus, \( \frac{\tan^{-1} n}{n^{1.1}} \leq \frac{\pi/2}{n^{1.1}} \). We want to compare this series to the simpler series \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \).

Step 3: Compare to a p-Series

The series \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \) is a constant multiple of the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \). A p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \). Here, \( p = 1.1 > 1 \), so \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) converges, and thus \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \) converges as well.

Step 4: Conclude Using the Comparison Test

Since \( \frac{\tan^{-1} n}{n^{1.1}} \leq \frac{\pi/2}{n^{1.1}} \) and \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \) is convergent, by the Comparison Test, the original series \( \sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1.1}} \) also converges.

Key Concepts

Comparison Test

The comparison test is a powerful tool for analyzing the convergence or divergence of an infinite series. This test involves comparing a series we are interested in with another series whose convergence properties are already known. If the series we are interested in is smaller in terms through an inequality with a known convergent series, then it also converges. Similarly, if it is larger than a known divergent series, it diverges.

For the series \( \sum_{n=1}^{\infty} \frac{\tan^{-1} n}{n^{1.1}} \), we use the comparison test by comparing it to \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \). Since \( \tan^{-1} n \) is bounded by \( \pi/2 \), the expression \( \frac{\tan^{-1} n}{n^{1.1}} \leq \frac{\pi/2}{n^{1.1}} \). Given that \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \) is a convergent series, our original series converges by this comparison.

Infinite Series

An infinite series is essentially a sum of infinitely many terms, written as \( \sum_{n=1}^{\infty} a_n \). These series can represent an array of sequences and appear in various mathematical applications, from calculus to complex analysis. Understanding whether an infinite series converges (sums to a finite limit) or diverges (sums to infinity or doesn't settle on a limit) is central to these applications.

For the series in the exercise, the infinite series is expressed as \( \sum_{n=1}^{\infty} \frac{\tan^{-1} n}{n^{1.1}} \). Identifying the methods such as the comparison test to determine its behavior (convergent or divergent) is crucial. The fact that this series converges helps us understand its long-term behavior, and assures us of reaching a finite sum regardless of its infinite nature.

p-Series

A pivotal concept when dealing with series is the p-series, signed as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). A p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). This simple yet powerful rule allows mathematicians to quickly determine the convergence or divergence of series modeled in this format.

In our exercise, \( \sum_{n=1}^{\infty} \frac{\pi/2}{n^{1.1}} \) is a constant multiple of the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \). Here, \( p = 1.1 \), thus confirming convergence since \( 1.1 > 1 \). Recognizing and leveraging this property of the p-series makes the analysis of more complex series, like the one initially given, much simpler when used in conjunction with a comparison strategy.

Bounded Functions

A bounded function is one whose outputs stay within a fixed range for all inputs in its domain. This characteristic is influential in the evaluation of series. The function \( \tan^{-1} n \) in our problem is bounded. Its values are restricted between 0 and \( \pi/2 \) for all natural \( n \).

Knowing that \( \tan^{-1} n \) is bounded simplifies comparisons. It allows us to substitute \( \tan^{-1} n \) with \( \pi/2 \) for inequalities, making problem-solving more straightforward. This bound was essential in establishing the inequality \( \frac{\tan^{-1} n}{n^{1.1}} \leq \frac{\pi/2}{n^{1.1}} \) during the comparison test. Thus, recognizing and applying the properties of bounded functions can significantly facilitate the convergence analysis of series like the one in the exercise.