Question

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} \tan ^{-1} k}{k^{3}}$$

Short answer

Question: Determine if the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k} \tan ^{-1} k}{k^{3}}\) converges absolutely, conditionally, or diverges. Answer: The series converges absolutely.

Step-by-step solution

Apply the Absolute Convergence Test

Let's begin by considering the absolute value of each term in the series: $$|a_n| = \left|\frac{(-1)^{k} \tan ^{-1} k}{k^{3}}\right| = \frac{\tan^{-1} k}{k^3}$$ We want to examine the convergence of $$\sum_{k=1}^{\infty} \frac{\tan^{-1} k}{k^3}$$ Now let's use the Comparison Test. The Comparison Test states that if \(0 \leq a_n \leq b_n\) for all \(n\) and the series \(\sum_{n=1}^{\infty} b_n\) converges, then \(\sum_{n=1}^{\infty} a_n\) converges as well. Since we know that \(\tan^{-1} k \leq \frac{\pi}{2}\) and \(k^3 \geq 1\) for all \(k\geq1\), we can establish that: $$0 \leq \frac{\tan^{-1} k}{k^3} \leq \frac{\pi}{2k^3}$$ Now, let's consider the series $$\sum_{k=1}^{\infty} \frac{\pi}{2k^3} = \frac{\pi}{2}\sum_{k=1}^{\infty} \frac{1}{k^3}$$ The last series is a p-series with \(p=3\), so it converges. Therefore, by the Comparison Test, the series \(\sum_{k=1}^{\infty} \frac{\tan^{-1} k}{k^3}\) also converges. As a result, the given series converges absolutely.

Conclusion

Since the series converges absolutely according to the Absolute Convergence Test, it is not necessary to apply the Alternating Series Test. Thus, we can conclude that the given series $$\sum_{k=1}^{\infty} \frac{(-1)^{k} \tan ^{-1} k}{k^{3}}$$ converges absolutely.

Key concepts

Absolute Convergence

Absolute convergence is a fundamental concept when evaluating the convergence of a series. For a series \( \sum a_n \), if the series formed by taking the absolute values of each term, \( \sum |a_n| \), converges, then the original series \( \sum a_n \) is said to converge absolutely. This is a stronger condition than regular convergence because it implies that the series retains its convergence even if all terms were positive.

Why is this so important? If a series converges absolutely, it indicates a certain robustness and stability in its behavior. It can withstand changes without becoming divergent. If you can show that \( \sum |a_n| \) converges, this gives you a powerful assurance about the original series' behavior.

In our exercise, the series involves terms that include the inverse tangent function, modified by a factor of \((-1)^k\). However, by examining the absolute values, the alternating sign is neutralized, making it easier to apply comparison tests and other convergence criteria.

Comparison Test

The Comparison Test is a useful technique for deciding the behavior of series convergence. It involves comparing the series in question with another series whose convergence properties are already known.

Here's how it works: if you have a series \( \sum a_n \) and can find another series \( \sum b_n \) where \( 0 \leq a_n \leq b_n \) for all sufficiently large \( n \), and \( \sum b_n \) is known to converge, then \( \sum a_n \) will also converge. If \( \sum b_n \) diverges, it leads to the conclusion that \( \sum a_n \) also diverges.

• *Converging Test:* If \( \sum b_n \) converges and \( a_n \leq b_n \), then \( \sum a_n \) converges.
• *Diverging Test:* If \( \sum b_n \) diverges and \( a_n \geq b_n \), then \( \sum a_n \) diverges.

In the given solution, the series \( \frac{\tan^{-1}k}{k^3} \) is compared to \( \frac{\pi}{2k^3} \), a known convergent series, which allowed us to conclude that the given series converges absolutely.

P-series

A p-series is a specific form of series written as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of a p-series is closely tied to the value of \( p \):

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

This characteristic makes p-series very handy reference points for comparison when examining other series for convergence.

In our exercise, the comparison was made with the series \( \frac{1}{k^3} \), where \( p = 3 \). Since this series converges due to having \( p > 1 \), it serves as a strong comparison for establishing that the target series also converges absolutely.

Alternating Series Test

The Alternating Series Test is particularly helpful when dealing with series that have alternating positive and negative terms. It provides a set of conditions under which an alternating series converges.

To apply the Alternating Series Test, two main conditions must be satisfied:

• The absolute value of the terms \( |a_n| \) must be decreasing: \(|a_{n+1}| \leq |a_n|\).
• The limit of the terms as \( n \) approaches infinity must be zero: \( \lim_{n \to \infty} a_n = 0 \).

For the given exercise, it wasn't necessary to use the Alternating Series Test because absolute convergence had already been established. Nevertheless, this test remains a valuable tool for exploring series that feature an alternating pattern but may not meet the criteria for absolute convergence.