Problem 19 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} n !(-e)^{-n} $$

Step-by-Step Solution

Answer

The series diverges by the n-th term test.

Step 1: Identify the Series and Test for Divergence

To determine if the series \( \sum_{n=1}^{\infty} n!(-e)^{-n} \) converges or diverges, first consider the terms of the series. Notice the general term of the series, \( a_n = \frac{n!}{(-e)^n} \). A good starting test for series convergence is the n-th term test for divergence, which states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum a_n \) diverges.

Step 2: Apply the n-th Term Test

Evaluate the limit \( \lim_{n \to \infty} \frac{n!}{(-e)^n} \). As \( n \to \infty \), the factorial function \( n! \) grows extremely fast compared to the exponential term \((-e)^n\), causing \( a_n \to \infty \). Hence, \( \lim_{n \to \infty} a_n eq 0 \).

Step 3: Conclusion Based on the n-th Term Test

Since \( \lim_{n \to \infty} \frac{n!}{(-e)^n} eq 0 \), by the n-th term test, the series \( \sum_{n=1}^{\infty} n!(-e)^{-n} \) diverges.

Key Concepts

n-th term test

The n-th term test is a fundamental tool in determining whether a series diverges. It focuses on the behavior of the individual terms of a series as the series progresses towards infinity. When you want to find out if a series diverges, you'll look at the limit of its terms.

Here's how the test works:

• Take the general term of the series, which is usually denoted as \( a_n \), and find its limit as \( n \) approaches infinity.
• If this limit \( \lim_{n \to \infty} a_n eq 0 \), the series diverges.
• If the limit equals zero, the test is inconclusive, which means the series could either converge or diverge, and other tests need to be applied.

In our specific exercise, we applied this test to the term \( \frac{n!}{(-e)^n} \). Since the limit does not equal zero, the series is confirmed to diverge.

factorial function

The factorial function, denoted by \( n! \), is a mathematical function that is essential in many areas of mathematics. It represents the product of all positive integers from 1 to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow very quickly as \( n \) increases, much faster than exponential and most other functions. This rapid growth is key to understanding why certain series with factorials like \( \sum_{n=1}^{\infty} \frac{n!}{(-e)^n} \) can diverge.

In this case, the factorial function \( n! \) increases extremely fast, which results in the individual terms of the series also increasing without bound. This behavior prevents the terms from approaching zero, leading to divergence, as shown by the n-th term test.

limit evaluation

Limit evaluation is a crucial technique in calculus used to understand the behavior of functions as the input approaches a certain value, often infinity. When dealing with series, especially in convergence tests, evaluating limits helps to determine the infinite trend of the terms.

For the series \( \sum_{n=1}^{\infty} \frac{n!}{(-e)^n} \), evaluating the limit of the general term \( \frac{n!}{(-e)^n} \) as \( n \to \infty \) was essential. Understanding that \( n! \) grows significantly faster than \( (-e)^n \) allowed us to confidently find that the limit tends to infinity.

Here are key points for evaluating limits:

• Recognize the relative growth rates of different functions: factorials outgrow exponentials as \( n \to \infty \).
• If the terms grow without bound, the limit evaluation suggests divergence.
• A careful comparison of term growth often reveals the overall behavior of the series.

divergence of series

Understanding the divergence of series is crucial for identifying series that do not converge to a definite value. In simple terms, a series diverges if the sum of its terms grows indefinitely or oscillates without approaching a specific limit.

The exercise in question has shown that the series \( \sum_{n=1}^{\infty} \frac{n!}{(-e)^n} \) diverges. This conclusion was reached because the limit of its terms is not zero, a key indicator of divergence as per the n-th term test.

There are a few reasons why a series might diverge:

• The terms do not approach zero, as required for potential convergence.
• The partial sums of the series increase indefinitely or vary wildly.
• Specific tests such as the n-th term test provide straightforward divergence evidence.

Recognizing divergence helps prevent miscalculations and misinterpretations of series behavior in mathematical and practical applications.