Problem 2 - Thomas Calculus

Question

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4} .\) $$ a_{n}=\frac{1}{n !} $$

Step-by-Step Solution

Answer

\( a_1 = 1 \), \( a_2 = \frac{1}{2} \), \( a_3 = \frac{1}{6} \), \( a_4 = \frac{1}{24} \).

Step 1: Understanding the Formula

The formula given is \( a_n = \frac{1}{n!} \), where \( a_n \) is the \( n \)th term of the sequence, and \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \).

Step 2: Calculating \( a_1 \)

Substitute \( n = 1 \) into the formula to find \( a_1 \): \[ a_1 = \frac{1}{1!} = \frac{1}{1} = 1 \].

Step 3: Calculating \( a_2 \)

Substitute \( n = 2 \) into the formula to find \( a_2 \): \[ a_2 = \frac{1}{2!} = \frac{1}{2 \times 1} = \frac{1}{2} \].

Step 4: Calculating \( a_3 \)

Substitute \( n = 3 \) into the formula to find \( a_3 \): \[ a_3 = \frac{1}{3!} = \frac{1}{3 \times 2 \times 1} = \frac{1}{6} \].

Step 5: Calculating \( a_4 \)

Substitute \( n = 4 \) into the formula to find \( a_4 \): \[ a_4 = \frac{1}{4!} = \frac{1}{4 \times 3 \times 2 \times 1} = \frac{1}{24} \].

Key Concepts

Sequences

A sequence is an ordered list of numbers that often follow a specific pattern or formula. Each number in the sequence is referred to as a "term." Sequences can be finite, with a limited number of terms, or infinite, with terms continuing indefinitely.

When working with sequences, it's essential to understand the rule or formula that generates the terms. For example, in the sequence given by the exercise, the terms are generated by the formula \( a_n = \frac{1}{n!} \). This means that each term is the reciprocal of the factorial of \( n \).

Some points to remember about sequences:

• They have specific positions or indices, like the first term, second term, and so on.
• The difference or ratio between consecutive terms can define certain types of sequences, like arithmetic or geometric sequences.
• Understanding the generating formula is key to finding any term in the sequence.

nth term

The \( n \)th term in a sequence is a general expression that defines the position \( n \) of a sequence's term. This term tells you how to calculate or determine any specific position's term without having to list all the preceding numbers.

For the provided exercise, the \( n \)th term is given as \( a_n = \frac{1}{n!} \). This formula enables you to calculate any desired term in the sequence by substituting the position \( n \) into the formula.

Calculating the \( n \)th term has several benefits:

• It simplifies finding any term directly without listing all prior terms.
• It provides insight into the behavior and properties of the sequence.
• It helps identify patterns, such as every sequence having a decreasing nature as the terms become smaller.

Understanding the concept of the \( n \)th term helps you unravel the sequence without confusion.

Factorial Formula

The factorial of a number \( n \) is a mathematical operation that multiplies the number by all preceding positive integers. It is represented with an exclamation mark, \( n! \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

The factorial formula is essential in sequences for determining terms like \( a_n = \frac{1}{n!} \). This indicates that each term in the sequence is the reciprocal of \( n! \). As \( n \) increases, \( n! \) grows significantly, which leads to the terms in the sequence becoming smaller.

Key characteristics of the factorial formula include:

• It is inherently linked with permutations and combinations, essential in probability and statistics.
• Factorials grow rapidly, contributing to very small fractions when creating sequences like \( a_n = \frac{1}{n!} \).
• Understanding how factorials work aids in solving problems involving series, combinatorics, and algorithms.

Knowing what factorials are and how they function is vital for delving deeply into mathematical sequences and other related areas.