Problem 43 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\begin{array}{l}{\sum_{n=2}^{\infty} \frac{1}{n !}} \\ {\text { (Hint: First show that }(1 / n !) \leq(1 / n(n-1)) \text { for } n \geq 2 . )}\end{array}\end{equation}

Step-by-Step Solution

Answer

The series \( \sum_{n=2}^{\infty} \frac{1}{n!} \) converges.

Step 1: Compare Terms with a Known Series

To determine convergence, compare the terms of the series \( \sum_{n=2}^{\infty} \frac{1}{n!} \) with the terms of another series that we know carefully. The hint suggests \( \frac{1}{n!} \leq \frac{1}{n(n-1)} \) for \( n \geq 2 \). This inequality is often used because \( \frac{1}{n(n-1)} \) is the typical term in a telescoping series.

Step 2: Analyze the Telescoping Series Comparison

Consider the telescoping series \( \sum_{n=2}^{\infty} \frac{1}{n(n-1)} \). This can be rewritten as partial fractions: \( \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} \). When expanded, the series forms a telescoping series, where all terms after the first cancel out. This implies the series converges since the sum simplifies to a finite number after factoring out the first terms.

Step 3: Apply Direct Comparison Test

Using the Direct Comparison Test, since \( 0 < \frac{1}{n!} \leq \frac{1}{n(n-1)} \) for \( n \geq 2 \) and \( \sum_{n=2}^{\infty} \frac{1}{n(n-1)} \) converges, by the test \( \sum_{n=2}^{\infty} \frac{1}{n!} \) also converges.

Key Concepts

Direct Comparison Test

The direct comparison test is a useful tool for determining the convergence or divergence of a series. This method involves comparing the series in question with another series whose convergence behavior is already known. Here’s how it works:

• If you can show that the terms of your series are less than or equal to the terms of a known converging series, then your series also converges.
• Conversely, if the terms are greater than or equal to the terms of a known diverging series, your series will diverge.

In this exercise, we apply the direct comparison test to the series \( \sum_{n=2}^{\infty} \frac{1}{n!} \). By comparing with the telescoping series mentioned, which converges, we show that \( \frac{1}{n!} \) is less than or equal to \( \frac{1}{n(n-1)} \) for \( n \geq 2 \). This satisfies the requirements of the test, leading us to conclude that the original series converges.

Telescoping Series

A telescoping series is a series where, after expansion, most terms cancel each other. This makes it easier to find the sum, as you're left with just a few terms. In the given exercise, the series \( \sum_{n=2}^{\infty} \frac{1}{n(n-1)} \) is compared to the original series.

The series can be rewritten using partial fractions:\[ \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} \]Expanding this into individual terms, you'll notice that intermediate terms cancel out, similar to closing the leaves of a telescope.

• This cancellation leaves only the first and the last terms to determine the sum.
• As most terms cancel each other out, the series converges to a finite value, as seen in the convergence of the telescoping series.

This characteristic is particularly helpful when using it in conjunction with the direct comparison test to show convergence for other series, as was done with \( \sum_{n=2}^{\infty} \frac{1}{n!} \).

Factorial Series

Factorial series involve mathematical expressions with factorials, such as \( \frac{1}{n!} \). A factorial, denoted \( n! \), is the product of an integer and all the positive integers below it. Factorials grow very quickly, much faster than exponential growth.

In series using factorials like \( \sum_{n=2}^{\infty} \frac{1}{n!} \):

• As \( n \) increases, \( n! \) rapidly increases, making \( \frac{1}{n!} \) decrease rapidly towards zero.
• Due to this rapid decrease, factorial series often converge because the terms diminish quickly.

When comparing factorial terms to simpler terms like \( \frac{1}{n(n-1)} \), we notice such factorial terms are indeed smaller, aiding us in applying the direct comparison test for convergence. This property of diminishing terms also supports the reason why factorial series lead to convergence.