Problem 43 - Thomas Calculus

Question

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{8 \tan ^{-1} n}{1+n^{2}} $$

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test.

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} \frac{8 \tan^{-1} n}{1+n^{2}} \). It resembles an infinite series and we need to determine whether it converges or diverges.

Step 2: Check for Convergence Using Comparison Test

The function \( \tan^{-1} n \) behaves similarly to \( \frac{\pi}{2} \) as \( n \to \infty \), so we approximate \( \frac{8 \tan^{-1} n}{1+n^{2}} \approx \frac{8 \cdot \frac{\pi}{2}}{n^2} = \frac{4\pi}{n^2} \). We suspect that the series is similar to the p-series \( \sum \frac{1}{n^2} \), which converges because \( p=2>1 \).

Step 3: Use Limit Comparison Test

Apply the Limit Comparison Test with \( b_n = \frac{4\pi}{n^2} \). Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{8\tan^{-1} n}{1+n^2} \cdot \frac{n^2}{4\pi} \). Since \( \tan^{-1} n \approx \frac{\pi}{2} \) for large \( n \), we get \( \lim_{n \to \infty} \frac{8\cdot \frac{\pi}{2}}{1+n^2} \cdot \frac{n^2}{4\pi} = \lim_{n \to \infty} \frac{n^2}{1+n^2} = 1 \). The limit is a finite nonzero constant, so by the Limit Comparison Test, the series converges.

Step 4: Conclusion on Convergence

Since the comparison series \( \sum \frac{4\pi}{n^2} \) converges, and the Limit Comparison Test yields a finite nonzero limit, we conclude that the original series \( \sum_{n=1}^{\infty} \frac{8 \tan^{-1} n}{1+n^{2}} \) converges as well.

Key Concepts

Comparison Test

The comparison test is a handy tool in evaluating the convergence of an infinite series. It involves comparing the series of interest to another series whose convergence behavior is known. For two series, \( \sum a_n \) and \( \sum b_n \), there are two parts to the test:

• If \( 0 \leq a_n \leq b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( 0 \leq b_n \leq a_n \) for all \( n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

This test is particularly useful when dealing with series that are complicated, as long as they can be bounded by more well-known series. In our case, we approximated the series \( \frac{8 \tan^{-1} n}{1+n^2} \) by comparison to the simpler series \( \frac{4\pi}{n^2} \), which is known to be convergent. Using the comparison test can save time and simplify the path to finding convergence.

Limit Comparison Test

The limit comparison test provides a less strict alternative to the regular comparison test. It allows for the comparison of two series without needing to find bounds for all terms. This test involves taking the limit of the ratio of the terms from the two series: \[\lim_{n \to \infty} \frac{a_n}{b_n} = L\]

• If \( L \gt 0 \) and finite, then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
• If \( L = 0 \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( L = \infty \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

For example, we used the limit comparison test by comparing our challenging series \( \frac{8\tan^{-1} n}{1+n^2} \) to the simpler known convergent series \( \frac{4\pi}{n^2} \). The limit turned out to be 1, a positive and finite number, confirming that the two series share the same convergence behavior.

Infinite Series

An infinite series is simply a sum of infinite terms. Expressed as \( \sum_{n=1}^{\infty} a_n \), it is a mathematical concept used to describe a sequence of numbers whose cumulative value might converge to a specific number or grow indefinitely. Understanding whether an infinite series converges or diverges is crucial in calculus and has many real-world applications, such as in physics and engineering.Typically, we want to find whether the total sum approaches a finite limit (converging) or does it keep increasing without bound (diverging). Techniques, such as the comparison test and limit comparison test, are vital to making these determinations.

P-Series

A p-series is a specific type of series that looks like \( \sum \frac{1}{n^p} \). The behavior of a p-series largely depends on the value of \( p \):

• When \( p \leq 1 \), the series diverges.
• When \( p \gt 1 \), the series converges.

The p-series serves as a fundamental benchmark when we're testing the convergence of other series. For instance, in the step solution, the original series was simplified to resemble \( \sum \frac{1}{n^2} \), which is a p-series with \( p = 2 \). Since this series converges, it helped us determine that the original series \( \sum \frac{8 \tan^{-1} n}{1+n^{2}} \) converges too, using comparison tests.