Problem 23 - Single Variable Calculus

Question

Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{n}}{n !}\)

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{n}}{n !}\) is not absolutely convergent, as shown by the Ratio Test, which resulted in a limit of \(e\). Furthermore, it is not conditionally convergent, as it does not meet the conditions of the Alternating Series Test. Therefore, the series is divergent.

Step 1: Determine Absolute Convergence

We first consider the absolute value of the series terms: \[\sum_{n=1}^{\infty} \left| \frac{(-1)^{n+1} n^{n}}{n !} \right| = \sum_{n=1}^{\infty} \frac{n^{n}}{n !}.\] To evaluate the convergence of this series, we can apply the Ratio Test. This test states that if \(\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L\), then the series \(\sum_{n=1}^{\infty} a_n\) is convergent if \(L < 1\) and divergent if \(L > 1\). If \(L = 1\), the test is inconclusive.

Step 2: Apply the Ratio Test for Absolute Convergence

Calculate the limit of the ratio of consecutive terms of the absolute value series: \[\lim_{n\to\infty} \left|\frac{ \frac{(n+1)^{n+1}}{(n+1) !} }{ \frac{n^{n}}{ n!} }\right| = \lim_{n\to\infty} \frac{(n+1)^{n+1} n!}{(n+1)! n^n}.\] Simplify the expression by canceling the factorials and noting that \((n+1)! = (n+1) n!\): \[\lim_{n\to\infty} \frac{(n+1)^n}{n^n}.\] Now, we can divide each term inside the parentheses \((n+1)^n\) by \(n^n\) and apply the limit: \[\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e.\] Since \(e > 1\), the Ratio Test indicates that the series \(\sum_{n=1}^{\infty} \frac{n^{n}}{n !}\) is divergent. Therefore, the original series is not absolutely convergent.

Step 3: Apply the Alternating Series Test

Since the series is not absolutely convergent, we will now check for conditional convergence using the Alternating Series Test. A series is conditionally convergent if it meets these two criteria: 1. The terms of the series are decreasing. 2. The limit of the terms approaches zero as \(n\) goes to infinity. Let's examine our original series: \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{n}}{n !}\). First, let's check if the terms of the series are decreasing: \[\frac{n^{n}}{n!} > \frac{(n+1)^{n+1}}{(n+1)!}.\] It is clear that the terms are indeed decreasing, since both the numerator and denominator are increasing, and the denominator is growing faster than the numerator. Now, let's check if the limit of the terms approaches zero as \(n\) goes to infinity: \[\lim_{n\to\infty} \frac{n^{n}}{n!}.\] Using the same argument from Step 2, we see that this limit converges to a constant value (non-zero), which is: \[\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n = e.\] Since the limit of the terms does not approach zero as \(n\) goes to infinity, the series fails the Alternating Series Test. This means that the series is not conditionally convergent.

Step 4: Conclusion

Since the series is neither absolutely convergent nor conditionally convergent, it is divergent.

Key Concepts

Absolute Convergence

A series is said to be absolutely convergent if the series of absolute values of its terms converges. For instance, if \ \( \sum_{n=1}^{\infty} a_n \) \ is a series, then taking the absolute value of each term gives us \ \( \sum_{n=1}^{\infty} |a_n| \). If this series converges, the original series is absolutely convergent.
In our exercise, we checked absolute convergence by evaluating the series \ \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n+1} n^{n}}{n !} \right| = \sum_{n=1}^{\infty} \frac{n^{n}}{n !} \).
The absolute convergence is crucial because it implies convergence regardless of the signs of the terms. Unfortunately, in this case, the absolute series was found to be divergent, meaning the original series cannot be absolutely convergent.

Ratio Test

The Ratio Test is a handy tool for determining the convergence of a series. It involves finding the limit of the ratio of consecutive terms. Mathematically, this is expressed as \ \( \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L \).

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

In our example, the limit was found to be \ \( e \), which is approximately 2.718, greater than 1. This resulted in the divergence of the series \ \( \sum_{n=1}^{\infty} \frac{n^{n}}{n !} \). Thus, the Ratio Test provided clear evidence that the series is divergent.

Alternating Series Test

The Alternating Series Test helps identify series that converge conditionally. A series alternates if its terms switch between positive and negative.
To apply this test, the series must:

• Have decreasing terms.
• Approach zero as \ \( n \) \ goes to infinity.

The original series \ \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{n}}{n !} \) has alternating signs. While the terms were decreasing, they did not approach zero. Instead, the limit was \ \( e \), a non-zero constant, failing the Alternating Series Test. Thus, the series wasn’t conditionally convergent.

Conditional Convergence

A series is conditionally convergent if it converges but doesn't converge absolutely. In simpler terms, the series without absolute values converges, yet the series with absolute values does not.
In our situation, both absolute convergence and conditional convergence were investigated. Since the absolute version of the series already diverged, conditional convergence became a question of whether the series can converge based solely on its alternating nature and decreasing terms. But the Alternating Series Test indicated failure to meet its criteria, ruling out conditional convergence.

Divergent Series

A divergent series is a series that doesn’t converge, meaning its terms don’t approach a specific limit. In mathematical terms, the partial sums of the series always grow without bound.
After applying both the Ratio Test and the Alternating Series Test to the given series, it was concluded that it didn’t satisfy the conditions for any type of convergence. Therefore, \ \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^{n}}{n !} \) is considered divergent.
This conclusion highlights the importance of applying different tests to understand the behavior of series comprehensively.