Problem 39 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series converges (not absolutely) by the Alternating Series Test.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty} (-1)^n \frac{(2n)!}{2^n n! n} \). This is an alternating series because it includes the term \((-1)^n\).

Step 2: Apply the Ratio Test

To determine absolute convergence, we apply the **Ratio Test** to the absolute values: \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]where \(a_n = \frac{(2n)!}{2^n n! n}\). Calculate:\[\frac{a_{n+1}}{a_n} = \frac{\frac{(2n+2)!}{2^{n+1} (n+1)! (n+1)}}{\frac{(2n)!}{2^n n! n}}\]Simplifying this expression for the ratio:

Step 3: Simplify the Ratio Test Expression

Substitute and simplify:\[\frac{(2n+2)!}{2^{n+1} (n+1)! (n+1)} \cdot \frac{2^n n! n}{(2n)!}\]The factorial expressions simplify greatly:\[\frac{(2n+2)(2n+1)(2n)!}{2 \cdot (n+1)n(n!) \, 2^n} \cdot \frac{2^n n!. n}{(2n)!}\]After cancelling terms and simplifying, compute the limit:

Step 4: Evaluate Limit for Ratio Test

Calculate the limit:\[\lim_{n \to \infty} \frac{4n^2 + 6n + 2}{2(n+1)n} = \lim_{n \to \infty} \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{2(1 + \frac{1}{n})}\]As \(n \to \infty\), this approaches \( \frac{4}{2} = 2 \).

Step 5: Decide the Convergence Based on Ratio Test

Since the result from the Ratio Test is \(2 > 1\), the series does not converge absolutely.

Step 6: Apply the Alternating Series Test

To determine if the series converges, use the **Alternating Series Test**. Check that \( b_n = \frac{(2n)!}{2^n n! n} \) is positive, decreasing, and \( \lim_{n \to \infty} b_n = 0 \).

Step 7: Evaluate Limit and Monotony for Alternating Series

Check if \( b_n \to 0 \) as \( n \to \infty \): Due to the factorial in the numerator and denominator, it can be shown that:\[\lim_{n \to \infty} \frac{(2n)!}{2^n n! n} = 0 \]Since the factorial in the denominator grows faster, \( b_n \) decreases.

Step 8: Conclusion on Convergence

The Alternating Series Test shows that the series \( \sum (-1)^n \frac{(2n)!}{2^n n! n} \) converges because \( b_n \to 0 \) and \( b_n \) is decreasing.

Key Concepts

Ratio Test

One of the common tests to determine whether a series converges absolutely is the Ratio Test. It is particularly useful for series with factorial terms or exponential elements. The intuitive idea is to examine how each term of the series compares to the next one as the series progresses.

• First, denote a term of the series by \(a_n\).
• Compute the limit of the absolute value of the ratio of successive terms as \(n\) approaches infinity: \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]
• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, it diverges.
• If the limit equals 1, the test is inconclusive.

In the given series, by applying the Ratio Test, we found the limit to be 2, which is greater than 1. This led to the conclusion that the series is not absolutely convergent. Remember, absolute convergence implies that the series converges regardless of whether the terms alternate in sign. Thus, it is a strong form of convergence, but not achieved here.

Alternating Series Test

The Alternating Series Test is another tool used to determine the convergence of a series, especially useful when series terms alternate in sign. A series has terms alternating in sign if its general form includes (-1) raised to a power involving n, for instance.

• Identify the sequence components \(b_n\), where the series is represented by \(\sum(-1)^n b_n\).
• Check if \(b_n\) is positive, indeed all terms need to consist of positive values.
• Ensure \(b_n\) is decreasing. This means each term should be less than the preceding term starting from some point.
• Finally, confirm that the limit as \(n\) approaches infinity for \(b_n\) is 0: \[\lim_{n \to \infty} b_n = 0 \]

In the context of the exercise, with \(b_n = \frac{(2n)!}{2^n n! n}\), each term checks these requirements as it decreases and tends towards zero over time. Therefore, the series is determined to converge by this test, even though it does not converge absolutely.

Factorials in Convergence Tests

Factorials, denoted by the symbol \(!\), are a fundamental part of mathematics, especially when applied in series tests. They increase very rapidly with larger numbers and often appear in series, as they are essential in describing permutations and combinations.

• In series like the one at hand, \( (2n)! \) in the numerator and \( n! \) in the denominator play crucial roles in determining convergence.
• Factorials in the numerator and denominator are often one of the primary reasons why a series might appear complex. They contribute significantly to the behavior of the series as \(n\) grows.
• The rapid growth of factorials influences limits in tests such as the Ratio Test and may help in showing trends in the Alternating Series Test.

In our exercise, factorials in both the numerator and denominator had to be simplified to manage terms correctly in the Ratio Test, revealing that \(b_n\), the absolute term from the series, decreases to zero. Understanding how factorials escalate or diminish permits a clearer insight into the series' convergence behavior.