Problem 37 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) 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 diverges by the Limit Comparison Test and Direct Comparison Test.

Step 1: Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty} \frac{8 \tan^{-1}(n)}{1+n^2} \). It resembles a series where each term is a function of \(n\), rather than a simple polynomial or exponential function. Hence, applying a convergence test suitable for such a function is appropriate here.

Step 2: Apply the Limit Comparison Test

The series \( \sum_{n=1}^{\infty} \frac{8 \tan^{-1}(n)}{1+n^2} \) can be estimated by comparing it with \( \sum_{n=1}^{\infty} \frac{1}{n} \). Observing the behavior of \( \tan^{-1}(n) \), for large \(n\), we know that \( \tan^{-1}(n) \to \frac{\pi}{2} \). Comparing \( \frac{\tan^{-1}(n)}{n^2} \) with \( \frac{1}{n} \), as \(n\to\infty\), shows that both approaches have similar behavior. But, using the test shows divergence because \( \sum \frac{1}{n} \) (harmonic series) diverges.

Step 3: Confirm with Direct Comparison Test

Each term \( \frac{8 \tan^{-1}(n)}{1+n^2} \) is larger than \( \frac{8}{1+n^2} \). We know \( \sum \frac{1}{n^2} \) converges, but \( \sum \frac{1+n^2}{n^2} \approx \sum \frac{1}{n} \). Since our original series' terms diminish at a rate slower than \( \sum \frac{8}{n^2} \) further confirms our divergence result since \( \sum \frac{8 \tan^{-1}(n)}{1} \gt \sum \frac{1}{n} \) for large \(n\), indicating convergence is implausible.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a handy tool when examining the convergence or divergence of a series. Especially those resembling other well-known series like the harmonic series.

Here's how it works in a nutshell:

• Given two series, \( \sum a_n \) and \( \sum b_n \), we consider the limit \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \).
• If the limit results in a positive number, say \( c \), where \( 0 < c < \infty \), then both series either converge or diverge simultaneously.

For the exercise provided, comparing the series \( \sum \frac{8 \tan^{-1}(n)}{1+n^2} \) with \( \sum \frac{1}{n} \) gives valuable insights. As \( n \to \infty \), \( \tan^{-1}(n) \approx \frac{\pi}{2} \), making \( \frac{\tan^{-1}(n)}{n^2} \to \frac{1}{n} \). The comparison led to the conclusion of divergence since the harmonic series \( \sum \frac{1}{n} \) is known to diverge.

Divergence

Divergence refers to when a series does not sum up to a finite number as the number of terms grows indefinitely. It's crucial to determine if a series converges or diverges, as it helps in understanding the solution behavior often seen in calculus.

When looking at the given series, the divergence was established using the Limit Comparison Test. By showing that the terms of our series behave similarly to the harmonic series, which does not converge, we conclude divergence.

• Divergence occurs when a series' sum grows indefinitely or oscillates without settling.
• If the test used, like the Limit Comparison or Direct Comparison, indicates that sum increases indefinitely, the series diverges.

Direct Comparison Test

The Direct Comparison Test is another significant method for determining the convergence or divergence of a series. This approach involves directly comparing the series term-by-term with another known series.

Steps to apply the Direct Comparison Test effectively:

• Find a series \( b_n \) which is easier to evaluate and is similar to the given series \( a_n \).
• For convergence: If \( 0 \leq a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• For divergence: If \( 0 \leq b_n \leq a_n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In our exercise, by comparing each term \( \frac{8 \tan^{-1}(n)}{1+n^2} \) with \( \frac{8}{1+n^2} \), we can observe that they decrease slower than a known convergent series \( \sum \frac{8}{n^2} \), reinforcing the divergent result from our limit comparison.