Question

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \frac{(-1)^{n} n !}{3^{n}} $$

Short answer

The series is divergent.

Step-by-step solution

Identify the Type of Series

The series given is \( \sum_{n=1}^{\infty} \frac{(-1)^{n} n!}{3^n} \). It is an alternating series because of the term \((-1)^n\), which means the signs of the terms alternate between positive and negative.

Apply the Alternating Series Test

To determine whether the series is convergent via the Alternating Series Test, check if \(a_n = \frac{n!}{3^n}\) (ignoring the alternating sign) satisfies the test conditions: (1) \(a_{n+1} \leq a_n\) eventually, and (2) \(\lim_{n \to \infty} a_n = 0\).

Evaluate the Limit of Terms

Calculate \(\lim_{n \to \infty} \frac{n!}{3^n}\). Since factorials \(n!\) grow much faster than exponential terms such as \(3^n\), \(\lim_{n \to \infty} \frac{n!}{3^n} = \infty\).

Conclusion on Convergence

Since \(\lim_{n \to \infty} a_n eq 0\), the alternating series test fails. Therefore, the series diverges.

Key concepts

Alternating Series Test

In mathematics, the Alternating Series Test is an essential tool to determine the convergence of an infinite series whose terms alternate in sign. An alternating series typically takes the form

• \(\sum_{n=1}^{\infty} (-1)^n a_n\), where \(a_n\) are positive terms.

To apply the Alternating Series Test, ensure these conditions:

• \(a_n\) is decreasing. This means that each term after the initial decreases in size. More formally, \(a_{n+1} \leq a_n\) should be true for some point onward.
• The limit of \(a_n\) as \(n\) approaches infinity is zero: \(\lim_{n \to \infty} a_n = 0\).

If both conditions are met, the series converges. If either is not met, as in our exercise, the series may diverge. In this problem, \(\lim_{n \to \infty} \frac{n!}{3^n} eq 0\), causing the alternating series test to fail, indicating potential divergence.

Factorial vs Exponential Growth

The comparison between factorial and exponential growth is crucial in understanding series convergence problems like the one in our exercise. Factorials are a product of integers from 1 to \(n\) and grow extremely fast. For example, \(6! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720\).

In contrast, exponential growth involves terms like \(b^n\) where \(b\) is a base raised to the power of \(n\). Even at large values of \(n\), factorial growth outpaces exponential growth.
In our series, \(a_n = \frac{n!}{3^n}\), as \(n\) increases, \(n!\) grows significantly faster compared to \(3^n\). This leads to the fraction increasing, ultimately diverging. This contrast explains why the alternating series test condition \(\lim_{n \to \infty} a_n = 0\) cannot be fulfilled.

Series Divergence

Understanding why a series diverges is important, as it tells us that the sum approaches infinity instead of settling to a finite number. Divergence happens when the terms of the series do not shrink to zero quickly enough or when they grow larger.

In the series from our exercise, \(\sum_{n=1}^{\infty} \frac{(-1)^{n} n!}{3^{n}}\), divergence is identified due to the failure of the Alternating Series Test. Since \(\lim_{n \to \infty} \frac{n!}{3^n} eq 0\), the terms \(a_n = \frac{n!}{3^n}\) do not become sufficiently small to ensure convergence.
Even though the terms' signs alternate, the overpowering growth of the factorial component causes the series sum to grow indefinitely.
Recognizing divergence helps us understand the behavior of infinite sums, which in many cases, is contrary to the expected potential convergence.