Problem 19 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} (-1)^n \frac {n^n}{n!} \)

Step-by-Step Solution

Answer

The series diverges by the Alternating Series Test.

Step 1: Identify the Series

We have the series \( \displaystyle \sum_{n = 1}^{\infty} (-1)^n \frac {n^n}{n!} \), which is an alternating series due to the term \((-1)^n\). To test its convergence, we will use the Alternating Series Test.

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that \( \sum (-1)^n a_n \) converges if (1) \( a_{n+1} \lt a_n \) for all \( n \) and (2) \( \lim_{n \to \infty} a_n = 0 \), where \( a_n = \frac{n^n}{n!} \). We will first evaluate the second condition, checking if \( \lim_{n \to \infty} \frac{n^n}{n!} = 0 \).

Step 3: Evaluate the Limit \( \lim_{n \to \infty} \frac{n^n}{n!} \)

Using Stirling's approximation \( n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \), we can approximate \( \frac{n^n}{n!} \approx \frac{n^n}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n} = \frac{e^n}{\sqrt{2\pi n}} \). As \( n \to \infty \), \( e^n \) grows exponentially, so \( \lim_{n \to \infty} \frac{e^n}{ \sqrt{2\pi n} } = \infty \). Thus, \( \lim_{n \to \infty} \frac{n^n}{n!} = \infty \) and the limit does not approach zero.

Step 4: Conclusion on Convergence

Since \( \lim_{n \to \infty} \frac{n^n}{n!} eq 0 \), the alternating series test fails. The series \( \sum_{n=1}^{\infty} (-1)^n \frac{n^n}{n!} \) diverges because the terms \( a_n \) do not tend towards zero as \( n \to \infty \).

Key Concepts

Alternating Series Test

The alternating series test is a handy tool to determine if a particular alternating series converges or diverges. An alternating series is characterized by a negative one raised to the power of the series index, typically indicating the alternation of signs. To apply the alternating series test, there are two simple conditions that need to be fulfilled:

• The absolute value of terms, called \( a_n \), must be decreasing. This means \( a_{n+1} < a_n \) for all \( n \).
• The limit of \( a_n \) as \( n \to \infty \) must be 0.

For the given series \( \sum_{n=1}^{\infty} (-1)^n \frac{n^n}{n!} \), it's clear that the first condition might seem a bit challenging to directly compute, so focusing on the second condition—the limit—is often more straightforward. However, if either of these conditions is not met, the series does not converge. In this example, we find out through the steps that the series does not meet the second condition, implying that it diverges.

Stirling's Approximation

Stirling's approximation is a principle used in mathematics to estimate factorials, especially for large numbers. It is stated as:\[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n\]This approximation simplifies complex expressions where factorials occur, allowing us to manage difficult series more easily. In the context of our series \( \sum_{n=1}^{\infty} (-1)^n \frac{n^n}{n!} \), Stirling's approximation helps evaluate the cumbersome factorial in the denominator. By substituting \( n! \) with this approximation, the term \( \frac{n^n}{n!} \) simplifies to approximately \( \frac{e^n}{\sqrt{2\pi n}} \). This approximation reveals critical insights into how the function behaves as \( n \) becomes very large.

Limit Evaluation

Evaluating limits is a fundamental aspect in determining the convergence or divergence of a series. In our series, we need to evaluate the limit:\[ \lim_{n \to \infty} \frac{n^n}{n!}\]After using Stirling’s approximation, the expression simplifies to \( \frac{e^n}{\sqrt{2\pi n}} \). This form makes it easier to see the behavior of the sequence as \( n \) increases. Notably, as \( n \to \infty \), the exponential growth of \( e^n \) outpaces the growth of the root \( \sqrt{2\pi n} \). Therefore, the limit does not approach zero but instead approaches infinity.Since the term \( a_n \) does not approach zero, the second criterion for the alternating series test is violated, confirming that the original series is divergent. Understanding how limits are evaluated with such approximations is key to analyzing complex series.