Problem 33 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{(-100)^{n}}{n !} $$

Step-by-Step Solution

Answer

The series converges absolutely by the ratio test.

Step 1: Identify the Series Type

The series given is \(\sum_{n=1}^{\infty} \frac{(-100)^n}{n!}\). This is a series where each term involves \((-100)^n\), which is an alternating sign, and \(n!\), which appears similar to a Taylor series expansion.

Step 2: Apply the Ratio Test

To determine absolute convergence, we use the ratio test. For a series \(\sum a_n\), the ratio test considers \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). If \(L < 1\), the series converges absolutely. Let's calculate:\[a_n = \frac{(-100)^n}{n!} \quad \text{and} \quad a_{n+1} = \frac{(-100)^{n+1}}{(n+1)!}\]\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{100^{n+1} / (n+1)!}{100^n / n!} = \frac{100}{n+1}\]So:\[L = \lim_{n \to \infty} \frac{100}{n+1} = 0\]Since \(L = 0 < 1\), the series converges absolutely.

Step 3: Conclusion for Convergence and Divergence

Since the series \(\sum \frac{(-100)^n}{n!}\) has \(L = 0\) with the ratio test, it is convergent and converges absolutely by the ratio test. A Vanishing limit of \(L\) assures that the factorial’s growth overcomes any effects of alternation or large base in exponent.

Key Concepts

Absolute Convergence

When dealing with series, understanding absolute convergence is crucial. A series is said to converge absolutely if the series formed by taking the absolute values of its terms also converges. This concept helps us determine the convergence behavior of a series, particularly those that contain alternating terms.

Here's why absolute convergence matters:

• Absolute convergence guarantees convergence, but not vice versa.
• If a series converges absolutely, then its terms can be rearranged in any order without affecting the sum.
• It simplifies analysis, especially when dealing with complex series.

In the context of our exercise, we used the ratio test to check for absolute convergence. By applying the absolute values in the test, we found that the limit was less than one, indicating that the series converges absolutely.

Factorial in Series

Factorials often appear in series, especially those resembling Taylor series or power series. The factorial, denoted by "!", represents the product of all positive integers less than or equal to a given number. For example, "5!" is equal to 5 x 4 x 3 x 2 x 1.

Factorials grow extremely quickly as the number increases, which impacts the convergence of a series. Understanding this growth pattern can help us determine the series’ behavior. In our specific series, each term included a factorial in the denominator. This rapid growth in the denominator is pivotal because:

• It tends to cause terms to become very small very quickly as n increases.
• It often overpowers any large constants in the numerator, as seen in our exercise with (-100)^n.

This property of factorials ensures that for sufficiently large values of n, the terms of the series approach zero, which aids in proving convergence.

Ratio Test

The ratio test is a powerful tool for determining the convergence or divergence of a series, particularly handy with series that show geometric or factorial-like growth. Here's how the ratio test works:

For a series \( \sum a_n \), we examine the limit: \[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]- If \(L < 1\), the series converges absolutely.- If \(L > 1\), or if the limit does not exist, the series diverges.- If \(L = 1\), the test is inconclusive.In the example exercise, the ratio of successive terms \( \left|\frac{a_{n+1}}{a_n} \right| = \frac{100}{n+1} \) resulted in a limit \(L = 0\), which is less than one. This informs us that the given series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \) converges absolutely. The factorial in the denominator contributed to making this result possible by quickly reducing term sizes compared to the constant 100 in the numerator.