Problem 8 - Thomas Calculus

Question

In Exercises \(1 - 14 ,\) determine if the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. $$ \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 10 ^ { n } } { ( n + 1 ) ! } $$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test.

Step 1: Identify the Series

The series given is \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 10 ^ { n } } { ( n + 1 ) ! } \). It is an alternating series because of the term \((-1)^n\), which causes the series to alternate in sign.

Step 2: State the Alternating Series Test

The Alternating Series Test states that a series \( \sum (-1)^n a_n \) converges if two conditions hold: 1. \( a_n \) is decreasing, 2. \( \lim_{{n \to \infty}} a_n = 0 \). Here, \( a_n = \frac{10^n}{(n+1)!} \).

Step 3: Check if \( a_n \) is Decreasing

To confirm that \( \frac{10^n}{(n+1)!} \) is decreasing, consider the ratio \( \frac{a_{n+1}}{a_n} = \frac{10^{n+1}/(n+2)!}{10^n/(n+1)!} = \frac{10}{n+2} \). For \( n \geq 1 \), \( \frac{10}{n+2} < 1 \), proving \( a_n \) is decreasing.

Step 4: Check the Limit of \( a_n \)

Let's find the limit of \( a_n \) as \( n \to \infty \): \[ \lim_{{n \to \infty}} \frac{10^n}{(n+1)!} \].Factorially grows faster than exponential terms, so as \( n \to \infty \), \( (n+1)! \to \infty \), making the limit 0: \[ \lim_{{n \to \infty}} \frac{10^n}{(n+1)!} = 0 \].

Step 5: Conclusion

Both conditions of the Alternating Series Test are satisfied: \( a_n \) is decreasing and \( \lim_{{n \to \infty}} a_n = 0 \). Thus, the series converges.

Key Concepts

Convergence of Series

The concept of series convergence refers to the behavior of a series as the number of terms increases indefinitely. A series converges when the sum approaches a specific finite number. For an alternating series, like the given exercise, the Alternating Series Test helps determine convergence. This test requires:

• The terms of the series, apart from the alternating sign factor, should decrease in magnitude.
• The limit of terms approaches zero as the series progresses.

Meeting these conditions ensures that the series doesn't veer off to infinity, but rather settles around a specific value. In mathematical problems, checking convergence is crucial since it signifies that adding infinitely many terms results in a manageable, finite sum.

Factorials in Series

Factorials, often represented by the symbol \(!\), play a pivotal role in series involving terms such as \( \frac{10^n}{(n+1)!} \). A factorial of a number \( n \) is the product of all positive integers up to \( n \). For instance, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).
In series analysis, factorials can determine the rate at which terms decrease. Since factorials grow extremely fast, they quickly become much larger than exponential terms. In our specific exercise, \((n+1)!\) grows rapidly with \(n\), causing the individual terms of the series to shrink significantly as \(n\) gets larger. This shrinking of terms is vital in the analysis of whether a series converges.

Limit of a Sequence

The limit of a sequence is a fundamental concept in calculus. It describes the value that the terms of a sequence approach as the index \( n \) becomes infinitely large. For any given sequence \( a_n \), if \( \lim_{{n \to \infty}} a_n = L \), then \( L \) is termed as the limit of the sequence.
In the Alternating Series Test, finding this limit helps verify one of the core conditions of convergence. If \( \lim_{{n \to \infty}} \frac{10^n}{(n+1)!} = 0 \), then the sequence terms become smaller indefinitely, supporting convergence. It’s like focusing on a shrinking balloon; as long as it shrinks to nothing, you're assured it won't go off endlessly.

Exponential vs Factorial Growth

Exponential and factorial growth both appear in mathematical expressions describing series, but they behave differently. Exponential growth is characterized by a term raised to the power of the series index, such as \( 10^n \). This means the term increases rapidly, but not nearly as quickly as factorial growth.
Factorial growth, denoted by \((n+1)!\) in the series, increases at an astonishing pace because it’s a product of sequential numbers. While \( 10^n \) quickly grows larger as \( n \) increases, \( (n+1)! \) overtakes it due to the powerful effect of multiplication inherent in factorials. In our series, the factorial term not only balances but overtakes exponential growth, leading to smaller terms and promoting series convergence.