Problem 33 - Thomas Calculus

Question

Which of the series in Exercises \(15 - 48\) 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

We are given the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \). This is an alternating series because of the factor \((-100)^n\), which alternates the sign of each term.

Step 2: Check Absolute Convergence

To check for absolute convergence, consider the series of absolute values, \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \). We will use the ratio test to determine if this series converges.

Step 3: Apply the Ratio Test

The ratio test involves looking at the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). For our series, we have \( a_n = \frac{100^n}{n!} \) and \( a_{n+1} = \frac{100^{n+1}}{(n+1)!} \). The ratio becomes:\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{100^{n+1} \cdot n!}{100^n \cdot (n+1)!} = \frac{100}{n+1} \] Then take the limit: \[ \lim_{n \to \infty} \frac{100}{n+1} = 0 \]

Step 4: Interpret the Ratio Test Result

Since the limit is \(0\), which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges. Therefore, the series of absolute values converges.

Step 5: Conclude Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges, our original series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \) converges absolutely.

Key Concepts

Alternating Series

An alternating series is a series whose terms alternate in sign. In the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \), the \((-100)^n\) factor causes each term to switch between positive and negative. This creates a back-and-forth effect in the progression of the series, which is characteristic of alternating series.

Alternating series can sometimes converge even when they do not converge absolutely. A key tool for testing the convergence of alternating series is the Alternating Series Test. This test states that an alternating series \( \sum (-1)^n a_n \) will converge if the following two conditions are met:

• The terms \( a_n \) are decreasing in absolute value, meaning \( |a_{n+1}| \leq |a_n| \).
• The limit of \( a_n \) as \( n \to \infty \) is zero: \( \lim_{n \to \infty} a_n = 0 \).

When these criteria are satisfied, the alternating series converges. However, in our analysis above, we are more broadly interested in absolute convergence.

Absolute Convergence

Absolute convergence refers to an infinite series \( \sum a_n \) that converges even when all its terms are replaced by their absolute values forming \( \sum |a_n| \). When a series converges absolutely, it means the corresponding non-alternating series \( \sum |a_n| \) also converges.

For example, in the series \( \sum_{n=1}^{\infty} \frac{(-100)^n}{n!} \), to check absolute convergence, we consider \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \). If this series converges, then our original series converges absolutely. Knowing this is helpful because absolute convergence assures convergence of series by straightforward manipulation of terms, contrary to some series that only satisfy conditional convergence.

Ratio Test

The Ratio Test is a popular method for testing convergence of series, especially those involving factorials or exponentials. It is particularly effective for series in the form \( \sum a_n \) where \( a_n \) involves terms like factorials or powers. The test assesses the limit of the absolute ratio of successive terms:
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

For the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \), the ratio becomes: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{100}{n+1} \]
Evaluating the limit gives:
\[ \lim_{n \to \infty} \frac{100}{n+1} = 0 \]
As the limit \( 0 \) is less than 1, the series \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) converges according to the Ratio Test. Therefore, the original series including the alternating factor converges absolutely.

Factorial Series

Factorials feature prominently in series when one factor of each term is expressed as \( n! \). A factorial series like \( \sum_{n=1}^{\infty} \frac{100^n}{n!} \) has terms that grow rapidly in size due to the factorial in the denominator \( n! \) quickly overpowering the constant raised to a power.

This typical quick growth of \( n! \) makes many such series converge. In our exercise, the presence of the factorial in \( n! \) ensures the terms of the unwanted portion of the ratio shrink dramatically as \( n \) increases, satisfying the Ratio Test.

• **Factorial Growth:** In the denominator, \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \), grows faster than any polynomial or exponential in the numerator.

Due to this fast growth, factorials dominate the convergence properties of the series they are part of, often leading to absolute convergence in series like ours.