Problem 38 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series converges absolutely by the ratio test.

Step 1: Analyze the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (n!)^2}{(2n)!} \). This is an alternating series due to the term \((-1)^{n+1}\).

Step 2: Absolute Convergence Check

To check for absolute convergence, we need to test if \( \sum_{n=1}^{\infty} \left| \frac{(n!)^2}{(2n)!} \right| \) converges. Since the absolute value of \((-1)^{n+1}\) is 1, the series we consider is \( \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} \).

Step 3: Use the Ratio Test for Absolute Convergence

We apply the ratio test to \( \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} \). Consider the limit \[ \lim_{n \to \infty} \left| \frac{\left((n+1)!\right)^2}{(2n+2)!} \div \frac{(n!)^2}{(2n)!} \right|.\] Simplifying this gives \[ \lim_{n \to \infty} \frac{(n+1)^2}{(2n+1)(2n+2)}.\]

Step 4: Evaluate the Limit for Ratio Test

Evaluating the limit, we get \( \lim_{n \to \infty} \frac{(n+1)^2}{(4n^2+6n+2)} = \lim_{n \to \infty} \frac{1}{4} = \frac{1}{4} \). Since \( \frac{1}{4} < 1 \), the series converges absolutely.

Key Concepts

Alternating Series

An alternating series is one where the terms alternate in sign, often because of a factor like \((-1)^{n+1}\). The key feature of alternating series is that they can converge even if their absolute values diverge. This is particularly because the terms flip between positive and negative, allowing them to potentially balance out.When analyzing an alternating series, we often utilize the **Alternating Series Test**. This test states that a series \(\sum (-1)^n a_n\) converges if:

• The absolute value of the terms \(|a_n|\) decreases steadily, i.e., \(a_{n+1} \leq a_n\) for all n.
• The limit of \(|a_n|\) as n approaches infinity is zero, i.e., \(\lim_{{n \to \infty}} a_n = 0\).

In the context of absolute convergence, however, we look beyond these alternating signs and check if the series converges without the alternating factor.

Absolute Convergence

Absolute convergence means that the series \(\sum a_n\) converges even when all terms are made positive, forming \(|a_n|\). If a series converges absolutely, it means it converges completely regardless of the term signs.To determine absolute convergence, we often take the absolute value of each term. If the resulting series converges, the original series is said to converge absolutely.In the exercise provided, the series originally includes the factor \((-1)^{n+1}\), which makes it an alternating series. By removing this sign-changing factor, we analyze the series:\[\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}\]If this absolute series converges through any test (like the ratio test), it implies that our original alternating series converges absolutely.

Ratio Test

The Ratio Test is a powerful method used to determine the convergence of a series. It is particularly useful for sequences involving factorials or powers. With the Ratio Test, we evaluate:The limit:\[L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|\]Where \(L\) will decide the behavior of the series:

• If \(L < 1\), the series converges.
• If \(L > 1\), the series diverges.
• If \(L = 1\), the test is inconclusive.

In our exercise, the Ratio Test is applied to the absolute series \(\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}\). Calculating:\[\lim_{n \to \infty} \frac{(n+1)^2}{(4n^2+6n+2)} = \frac{1}{4}\]Since \(rac{1}{4} < 1\), the series converges absolutely.

Factorials in Series

Factorials often appear in series, providing unique challenges and opportunities for analysis. A factorial, expressed as \(n!\), is the product of all positive integers less than or equal to \(n\). Factorials grow extremely fast as \(n\) increases, which can greatly influence the convergence of a series.In the series under consideration, both the numerator and the denominator involve factorials:\[(n!)^2 \, \text{and} \, (2n)!\]The presence of \(n!\) in the numerator and \( (2n)!\) in the denominator often results in rapidly decreasing terms, especially for large \(n\), due to the fact that \( (2n)!\) grows significantly faster than \(n!\).Understanding the behavior of factorials in series helps in identifying appropriate tests for convergence, such as the Ratio Test, which works efficiently with these forms because of the way relative growth rates affect the series' terms.