Problem 40 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(n !)^{2} 3^{n}}{(2 n+1) !} $$

Step-by-Step Solution

Answer

The series converges absolutely by the Ratio Test.

Step 1: Analyze the Series for Absolute Convergence

To determine if the series converges absolutely, we first consider the absolute value of the terms in the series: \[ a_n = \left| (-1)^n \frac{(n!)^2 3^n}{(2n+1)!} \right| = \frac{(n!)^2 3^n}{(2n+1)!}. \] We need to check if \( \sum_{n=1}^{\infty} a_n \) converges.

Step 2: Apply the Ratio Test for Absolute Convergence

The Ratio Test involves examining the limit:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]Calculate \( \frac{a_{n+1}}{a_n} \):\[ a_{n+1} = \frac{((n+1)!)^2 3^{n+1}}{(2(n+1)+1)!} = \frac{((n+1)!)^2 3^{n+1}}{(2n+3)!}. \]Thus,\[ \frac{a_{n+1}}{a_n} = \frac{((n+1)!)^2 3^{n+1} (2n+1)!}{(n!)^2 3^n (2n+3)!} = \frac{(n+1)^2 3}{(2n+3)(2n+2)}. \] Then,\[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)^2 3}{(2n+3)(2n+2)} = \lim_{n \to \infty} \frac{3(n+1)^2}{4n^2} = \frac{3}{4}. \] Since this limit is less than 1, the series converges absolutely.

Step 3: Conclusion on Convergence

Since the absolute series converges by the Ratio Test, the original series converges absolutely.

Key Concepts

Series Convergence

Understanding whether a series converges is a crucial step in analyzing how a series behaves as more terms are added. Let's break this down in a simple way:

• When we say a series converges, we're saying that as we add more and more terms of the series, the sum approaches a specific value. In mathematical terms, the sequence of partial sums approaches a finite number.
• On the other hand, if a series diverges, it does not settle towards any particular number. Instead, the sum could grow indefinitely or oscillate without approaching a fixed point.

There are different types of convergence:

• Absolute Convergence: A series is said to converge absolutely if the series of absolute values of its terms also converges.
• Conditional Convergence: A series converges conditionally if the series itself converges but the series of absolute values diverges.

Absolute convergence is a stronger condition than simple convergence because if a series converges absolutely, it converges unconditionally, independent of the order of its terms. In the problem's solution, we determined absolute convergence of a series using the Ratio Test.

Ratio Test

The Ratio Test is a popular method used to determine the convergence of an infinite series, especially for series with factorials, exponentials, or any series whose calculations heavily depend on the factors of consecutive terms.Here's how the Ratio Test works:

• Given a series with terms labeled as \(a_n\), the Ratio Test examines the limit of the absolute value of the ratio of consecutive terms as \(n\) approaches infinity. Specifically, this is analyzed with \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
• If this limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• If the limit is equal to 1, the test is inconclusive, meaning the series may converge or diverge, and another method must be used to determine its behavior.

In the exercise, we applied the Ratio Test to the absolute values of the series terms, and calculated the ratio of successive terms. The result was a limit of \(\frac{3}{4}\), which is less than 1, confirming absolute convergence.

Alternating Series

An alternating series is identified by its terms alternating in sign. For example, the series \((-1)^n a_n\), where each consecutive term is multiplied by \(-1\), thus switching signs.Key points about alternating series:

• Alternating Series Test: This test checks whether an alternating series converges by examining a few conditions. If the absolute value of the terms \(a_n\) decreases monotonically (each term is smaller than its predecessor) and approaches zero as \(n\) approaches infinity, then the series converges.
• However, even when an alternating series satisfies these conditions and converges, this doesn't automatically imply absolute convergence.

In the exercise given, the series is alternating because it includes a factor of \((-1)^n\). However, while tackling the problem, we focused on determining absolute convergence using other methods, like the Ratio Test. Alternating Series Test could also validate conditional convergence, but absolute convergence still holds more significance.