Problem 29 - Thomas Calculus

Question

Which of the series in Exercises \(15 - 48\) converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { \tan ^ { - 1 } n } { n ^ { 2 } + 1 } $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Analyze Series Type

The given series is \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { \tan ^ { - 1 } n } { n ^ { 2 } + 1 } \). This series is alternating due to the presence of \((-1)^n\), which alternates the signs of the terms.

Step 2: Check for Absolute Convergence

To determine absolute convergence, consider the series of absolute values: \( \sum _ { n = 1 } ^ { \infty } \left| \frac { \tan ^ { - 1 } n } { n ^ { 2 } + 1 } \right| \). Since \(\tan^{-1} n\) is an increasing function bounded between 0 and \(\pi/2\), and \(n^2 + 1\) grows without bounds, we need to compare this to a known convergent series. Since \(\frac { \tan^{-1} n }{ n^2 +1 } \approx \frac{\pi/2}{n^2}\), which resembles a \(p\)-series with \(p > 1\), the series of absolute values converges.

Step 3: Apply Alternating Series Test

To apply the Alternating Series Test, confirm that \(a_n = \frac{\tan^{-1} n}{n^2 + 1}\) satisfies the conditions: a) the terms \(a_n\) decrease in magnitude and b) \(\lim_{n \to \infty} a_n = 0\). Since \(\frac{\tan^{-1} n}{n^2 + 1} \to 0 \) as \(n \to \infty\) and the sequence is decreasing because both \(\tan^{-1} n\) grows slowly and \(n^2 + 1\) increases quickly, the series converges by the Alternating Series Test.

Key Concepts

Alternating Series Test

An alternating series is one where the signs of the terms alternate between positive and negative. The series \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } a_n \) is a classic example. To determine whether an alternating series converges, we use the Alternating Series Test. To apply this test, two main conditions must be satisfied:

• Firstly, the absolute values of the terms \( a_n \) need to decrease steadily. This means each term is smaller than the previous one, which ensures the terms are becoming negligible.
• Secondly, the limit of \( a_n \) as \( n \to \infty \) must be zero. This ensures that the tail-end of the series doesn't contribute significantly to the sum.

In our exercise, the given series alternates due to \((-1)^n\). As \( a_n = \frac{\tan^{-1} n}{n^2 + 1} \), and since the terms decrease and approach zero, the series passes the Alternating Series Test, confirming convergence.

Absolute Convergence

Absolute convergence is a stringent form of convergence. A series \( \sum a_n \) converges absolutely if the series of its absolute values \( \sum |a_n| \) also converges. Absolute convergence implies regular convergence, but the converse is not always true.To determine absolute convergence, one helpful approach is to consider series estimation and apply known convergence tests.In our example, we take the series \( \sum \left| \frac{\tan^{-1} n}{n^2 + 1} \right| \), which simplifies to considering the behavior of the terms \(\frac{\pi/2}{n^2+1}\) as \( n \to \infty \). This series resembles a \(p\)-series with \(p > 1\), ensuring that it converges by comparison with a known convergent series. Hence, the series converges absolutely.

P-Series

P-Series are special types of series that take the form \( \sum _ { n = 1 } ^ { \infty } \frac{1}{n^p} \), where \(p\) is a positive constant. The convergence behavior of a \(p\)-series is straightforward:

• If \(p > 1\), the series \( \sum \frac{1}{n^p} \) converges, due to the terms decreasing rapidly enough to have a finite sum.
• If \(0 < p \le 1\), the series diverges.

Understanding \(p\)-series is crucial in the comparison test for series.In our exercise, recognizing that the term \(\frac{\pi/2}{n^2}\) (similar to our transformed series) indicates convergence gives insight that \( n^2 + 1 \) grows similarly to a \(p\)-series with \(p = 2\), which converges since \(p > 1\).

Comparison Test

The Comparison Test is a valuable tool to determine the convergence or divergence of a series by comparing it to a known series. This test works by matching a series against another series whose convergence behavior is already understood.To apply the Comparison Test, you must:

• Find a series with positive terms that you know converges or diverges.
• Compare the terms of your series to the terms of the known series.
• If the terms of your series are less than or equal to the terms of the convergent series, then your series converges. Conversely, if they're greater than or equal to the terms of a divergent series, then your series diverges.

In the given exercise, to show absolute convergence, we compare \( \left| \frac{\tan^{-1} n}{n^2 + 1} \right| \) with a \(p\)-series of the form \(\frac{1}{n^2}\). Since \(\tan^{-1} n\) is bounded, this comparison is apt and aids in demonstrating convergence through known series behavior, establishing the absolute convergence of our series.