Problem 38 - Thomas Calculus

Question

Which of the series 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: Recognize the Type of Series

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

Step 2: Apply the Alternating Series Test

For an alternating series \( \sum (-1)^n a_n \), it converges if \( a_n \to 0 \) as \( n \to \infty \). Here, \( a_n = \frac{(n ! )^{2}}{(2 n)!} \). We need to check if \( a_n \to 0 \).

Step 3: Simplify the Term \(a_n\)

Calculate an approximation or bound of \( a_n = \frac{(n ! )^{2}}{(2 n)!} \). Note that \( (2n)! \) grows much faster than \( (n!)^2 \), suggesting that \( a_n \to 0 \) as \( n \to \infty \). Therefore, the series satisfies the Alternating Series Test condition.

Step 4: Apply the Ratio Test for Absolute Convergence

To test absolute convergence, apply the Ratio Test to \( \left| \frac{(n ! )^{2}}{(2 n)!} \right| \). Compute \( \lim_{n \to \infty} \frac{(n+1)^{2}}{(2n+2)!} \div \frac{n^{2}}{(2n)!} \), which simplifies to \( \frac{1}{4} \) using factorial properties and simplifications. Since \( \frac{1}{4} < 1 \), the series converges absolutely.

Step 5: Conclude the Convergence

Since the series converges absolutely according to the Ratio Test, it implies that the series also converges.

Key Concepts

Alternating Series Test

The alternating series test helps determine the convergence of series where the terms alternate in sign. For a series of the form \( \sum_{n=1}^{\infty} (-1)^n a_n \), two conditions must be met for the series to converge:

• The absolute value of the terms \( a_n \) must decrease steadily towards zero, i.e., \( a_{n+1} \leq a_n \).
• The limit of the terms must approach zero as \( n \) approaches infinity, i.e., \( a_n \to 0 \).

In our exercise, the term \( a_n \) is \( \frac{(n!)^2}{(2n)!} \). We observe that \( (2n)! \) grows faster than \( (n!)^2 \), which suggests \( a_n \to 0 \) as \( n \to \infty \). Thus, the series passes the alternating series test and therefore converges.

Absolute Convergence

Absolute convergence is a stronger form of convergence. If a series converges absolutely, it implies convergence of the series itself. A series \( \sum a_n \) is said to converge absolutely if the series of absolute values \( \sum |a_n| \) converges.
To determine absolute convergence for our series, we first ignore the sign alternation and consider the series \( \sum \left| \frac{(n!)^2}{(2n)!} \right| \).

For our exercise, using the ratio test (explained further below), we find that the series converges absolutely because the limit of the ratio of successive terms is less than 1. This means that \( \sum \frac{(n!)^2}{(2n)!} \) converges absolutely, and hence, the original alternating series also converges absolutely.

Ratio Test

The ratio test is a powerful tool for determining the absolute convergence of a series. Given a series \( \sum a_n \), we consider the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \). If the limit of this ratio as \( n \) goes to infinity is less than 1, i.e.,

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1 \),

then the series converges absolutely.
In our exercise, applying the ratio test to \( a_n = \frac{(n!)^2}{(2n)!} \), we compute the limit
\[ \lim_{n \to \infty} \frac{(n+1)!^2}{(2n+2)!} \div \frac{n!^2}{(2n)!} = \lim_{n \to \infty} \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{1}{4} \]

• Since \( \frac{1}{4} < 1 \), our given series converges absolutely, confirming that the ratio test confirms absolute convergence.