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Magazine Chapter 11: Problem 29
Which of the series in Exercises 1鈥�36 converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\tan ^{-1} n}{n^{1.1}} $$
Short Answer
Expert verified
The series converges by the Comparison Test.
Step by step solution
01
Understanding the Problem
We need to determine whether the infinite series \( \sum_{n=1}^{\infty} \frac{\tan^{-1} n}{n^{1.1}} \) converges or diverges. We'll explore appropriate convergence tests to make this determination.
02
Recall the Comparison Test Basics
The Comparison Test is useful for series comparison. If \( 0 \le a_n \le b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges. Similarly, if \( \sum b_n \) diverges, and \( a_n \ge b_n \), then \( \sum a_n \) diverges.
03
Estimate the Behavior of \(a_n\)
The term of the series is \( a_n = \frac{\tan^{-1} n}{n^{1.1}} \). Note that \( \tan^{-1} n \) tends to \( \frac{\pi}{2} \) as \( n \to \infty \). So for large \( n \), \( a_n \approx \frac{\pi/2}{n^{1.1}} \).
04
Choose a Comparison Series
Compare \( a_n \approx \frac{\pi/2}{n^{1.1}} \) with \( b_n = \frac{1}{n^{1.1}} \). The series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) is a p-series with \( p = 1.1 > 1 \) and is known to converge.
05
Apply the Comparison Test
Since \( 0 \le \frac{\tan^{-1} n}{n^{1.1}} \le \frac{\frac{\pi}{2}}{n^{1.1}} \), and the series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.1}} \) converges, by the Comparison Test, the series \( \sum_{n=1}^{\infty} \frac{\tan^{-1} n}{n^{1.1}} \) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Comparison Test
The Comparison Test is a handy tool used in determining the convergence or divergence of infinite series. It works by comparing a series of interest with another series whose behavior is known.
To apply this test, two key conditions must be satisfied:
To apply this test, two key conditions must be satisfied:
- If you have two series \( \sum a_n \) and \( \sum b_n \) where \( 0 \le a_n \le b_n \) for all \( n \), and \( \sum b_n \) converges, then the series \( \sum a_n \) must also converge.
- Conversely, if \( \sum b_n \) diverges and \( a_n \ge b_n \) for all \( n \), then \( \sum a_n \) also diverges.
Infinite Series
In mathematical terms, an infinite series is the sum of the terms of an infinite sequence. It's represented as \( \sum_{n=1}^{\infty} a_n \), where each \( a_n \) is a term in the sequence.
The challenge with infinite series is determining whether they converge (approach a finite limit) or diverge (grow without bound).
The challenge with infinite series is determining whether they converge (approach a finite limit) or diverge (grow without bound).
- A series converges if the sum of its terms approaches a specific value as more terms are added.
- On the other hand, a series diverges if it does not approach a specific value. This means the sum either becomes infinitely large or oscillates without settling to a limit.
P-Series
The p-series is a special type of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of a p-series is determined by the value of \( p \).
There are some pivotal rules regarding p-series convergence:
There are some pivotal rules regarding p-series convergence:
- If \( p > 1 \), the series converges. This is because the terms \( \frac{1}{n^p} \) become smaller at a rate fast enough that the sum approaches a finite limit.
- If \( p \le 1 \), the series diverges. In this case, the terms do not decrease rapidly enough to bring the sum to a finite limit.
Arctangent Function
The arctangent function, denoted as \( \tan^{-1} x \), is the inverse of the tangent function. It maps each real number \( x \) to an angle whose tangent is \( x \).
As \( x \) approaches infinity, \( \tan^{-1} x \) approaches \( \frac{\pi}{2} \). This limiting behavior is crucial in estimating the terms of a series involving \( \tan^{-1} x \).
As \( x \) approaches infinity, \( \tan^{-1} x \) approaches \( \frac{\pi}{2} \). This limiting behavior is crucial in estimating the terms of a series involving \( \tan^{-1} x \).
- In the context of series, as in the exercise, \( \tan^{-1} n \) behaves predictably for large \( n \), which informs approximations such as \( \frac{\tan^{-1} n}{n^{1.1}} \approx \frac{\pi/2}{n^{1.1}} \).
