Problem 8 - Calculus: Early Transcendentals

Question

List the first five terms of the sequence. \( a_n = \frac {(-1)^nn}{n! + 1} \)

Step-by-Step Solution

Answer

The first five terms are \(-\frac{1}{2}, \frac{2}{3}, -\frac{3}{7}, \frac{4}{25}, -\frac{5}{121}\).

Step 1: Understand the Sequence Formula

The sequence is defined by the formula \( a_n = \frac{(-1)^n n}{n! + 1} \). Here, \( (-1)^n \) determines the sign of the term depending on whether \( n \) is odd or even. \( n! \) represents factorial of \( n \), which is the product of all positive integers up to \( n \). The term \( n! + 1 \) is the denominator.

Step 2: Compute the First Term (n=1)

Substitute \( n = 1 \) into the formula: \( a_1 = \frac{(-1)^1 \times 1}{1! + 1} = \frac{-1 \times 1}{1 + 1} = \frac{-1}{2} \).

Step 3: Compute the Second Term (n=2)

Substitute \( n = 2 \) into the formula: \( a_2 = \frac{(-1)^2 \times 2}{2! + 1} = \frac{1 \times 2}{2 + 1} = \frac{2}{3} \).

Step 4: Compute the Third Term (n=3)

Substitute \( n = 3 \) into the formula: \( a_3 = \frac{(-1)^3 \times 3}{3! + 1} = \frac{-1 \times 3}{6 + 1} = \frac{-3}{7} \).

Step 5: Compute the Fourth Term (n=4)

Substitute \( n = 4 \) into the formula: \( a_4 = \frac{(-1)^4 \times 4}{4! + 1} = \frac{1 \times 4}{24 + 1} = \frac{4}{25} \).

Step 6: Compute the Fifth Term (n=5)

Substitute \( n = 5 \) into the formula: \( a_5 = \frac{(-1)^5 \times 5}{5! + 1} = \frac{-1 \times 5}{120 + 1} = \frac{-5}{121} \).

Key Concepts

Factorial

In mathematics, a factorial is a function that multiplies a number by every number below it. It is expressed with an exclamation mark. So, the factorial of a positive integer \( n \) is written as \( n! \). This means: \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \). For example:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

The factorial is crucial in permutations and combinations, and it often appears in sequences and series. It grows very quickly with increasing numbers, making it both a fundamental and intriguing part of mathematics.

Alternating Series

An alternating series is a sequence where the signs of the terms alternate between positive and negative. The sequence is often related to the term \((-1)^n\), which helps define whether the result is positive or negative:

• If \( n \) is even, \((-1)^n = 1\) and the term is positive.
• If \( n \) is odd, \((-1)^n = -1\) and the term is negative.

This alternation is evident in the sequence given by the formula \( a_n = \frac{(-1)^n n}{n! + 1} \). As \( n \) increases, the pattern of signs (+, -, +, -, ...) provides a balance, which leads to improved convergence properties when summing the series. The drastic changes in sign have interesting implications for the overall behavior of the sequence or series.

Sequence Terms

In any sequence, each number is referred to as a term. The terms follow a specific order and are generally defined by a formula, such as \( a_n = \frac{(-1)^n n}{n! + 1} \) in our exercise. Here are the first five terms calculated:

• \( n = 1 \): \( a_1 = \frac{-1}{2} \)
• \( n = 2 \): \( a_2 = \frac{2}{3} \)
• \( n = 3 \): \( a_3 = \frac{-3}{7} \)
• \( n = 4 \): \( a_4 = \frac{4}{25} \)
• \( n = 5 \): \( a_5 = \frac{-5}{121} \)

Each term is distinct and calculated by plugging \( n \) into the sequence formula. This allows students to study the behavior, convergence, or divergence of the sequence, as well as the influence of each component within the formula, like \( (-1)^n \), \( n \), and \( n! + 1 \). Understanding sequence terms is fundamental in learning progression throughout mathematics.

Mathematical Formula

Mathematical formulas are recipes for working out numbers. They are concise and precise, saving time and reducing errors. The formula \( a_n = \frac{(-1)^n n}{n! + 1} \) combines multiple mathematical concepts:

• **Alternation** via \( (-1)^n \)
• **Progressive terms** with \( n \)
• **Factorials** as \( n! + 1 \)

Each part of the formula plays a role in forming the sequence. Alternation causes some terms to be positive and some negative. Factorials grow rapidly, affecting the denominator and magnitude of each fraction, thus influencing how quickly these terms approach zero. Mastery of mathematical formulas like these is a key step in advancing in calculus and can significantly help in solving more complex mathematical problems.