Question

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{n^{n}} $$

Short answer

The series is conditionally convergent.

Step-by-step solution

Determine Type of Series

First, identify that this series \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{n^{n}} \) is an alternating series because of the \((-1)^n\) term, which alternates the sign of each term.

Apply the Alternating Series Test

The Alternating Series Test states that an alternating series \( \sum (-1)^n a_n \) converges if two conditions are met: \( a_n \) is positive, decreasing, and \( \lim_{n \to \infty} a_n = 0 \). Here, \( a_n = \frac{n!}{n^n} \).

Check if Terms are Decreasing

We need to check if \( a_n = \frac{n!}{n^n} \) is decreasing. To do this, compare \( a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}} \) with \( a_n \):\If \( a_{n+1} < a_n \), then the series is decreasing. Simplifying \( \frac{(n+1)!}{(n+1)^{n+1}} < \frac{n!}{n^n} \), we determine it is decreasing for large \( n \), as \( \frac{n}{n+1} \) becomes smaller than 1.

Compute the Limit of a_n

Find \( \lim_{n \to \infty} a_n \). Since factorial growth \( n! \) is outpaced by \( n^n \) (exponential growth), \( \lim_{n \to \infty} \frac{n!}{n^n} = 0 \). This satisfies the second condition of the Alternating Series Test.

Apply Absolute Convergence Test

To check for absolute convergence, consider the series \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n} n!}{n^n} \right| = \sum_{n=1}^{\infty} \frac{n!}{n^n} \). We estimate \( \frac{n!}{n^n} \) as seen in the Ratio Test, which shows that this series diverges due to the rapid growth of \( n! \).

Conclusion: Series Convergence Type

The original series is convergent by the Alternating Series Test, but not absolutely convergent as its absolute counterpart diverges.

Key concepts

Alternating Series Test

The Alternating Series Test is a valuable tool for determining the convergence of a specific type of series—those where terms alternate in sign. For a series of the form - \( \sum (-1)^n a_n \)- it can potentially converge if two conditions are fulfilled: 1. The terms \( a_n \) are positive and decrease steadily 2. The limit of \( a_n \) as \( n \) approaches infinity is zero: \( \lim_{n \to \infty} a_n = 0 \)

In our exercise, the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{n^{n}} \) is alternating due to the \((-1)^n\) term. As calculated, the terms \( a_n = \frac{n!}{n^n} \) exhibit both the decreasing nature and the limit requirement, ensuring the series converges. Still, the test doesn't guarantee absolute convergence but confirms conditional convergence for alternating series.

Absolute Convergence

Absolute convergence is a stricter form of convergence. It requires that the series formed by taking the absolute values of its terms also converges. This concept is crucial because an absolutely convergent series will always be convergent, but not all convergent series are absolutely convergent.

For absolute convergence, we examine the series:- \( \sum_{n=1}^{\infty} \frac{n!}{n^n} \)This involves analyzing the behavior of the positive series without considering the alternating sign. However, this series diverges, indicating that our original series is not absolutely convergent. This result is important as it defines the original series only as conditionally convergent, due to its alternating nature.

Factorial Growth

Factorial growth describes the rapid increase in the values of factorial expressions \( n! \) as \( n \) gets larger. Compared to other types of growth, such as linear or polynomial, factorial growth is significantly faster.

When comparing factorial growth with exponential growth, like \( n^n \)—which appears in our series \( \frac{n!}{n^n} \)—factorial growth is outpaced. This relationship is useful in our analysis because it helps show that the sequence of terms \( a_n = \frac{n!}{n^n} \) decreases as \( n \) increases, thereby meeting the conditions of the Alternating Series Test.

Understanding these differences helps us predict which component of our expressions will dominate as \( n \) approaches infinity.

Convergence and Divergence Analysis

In analyzing series, understanding the distinction between convergence and divergence is essential. Convergence indicates that the series approaches a finite limit, while divergence means it does not.

Using the Alternating Series Test, we have already established convergence for our example series \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{n^{n}} \). However, via the Absolute Convergence Test, we determined that its absolute counterpart diverges:- \( \sum_{n=1}^{\infty} \frac{n!}{n^n} \)

Such divergence in the absolute series is crucial in identifying that the convergence of the original series is conditional. It's a subtlety in series analysis that helps differentiate between types of convergence and divergent behavior.