Problem 27 - Thomas Calculus

Question

Find a formula for the \(n\)th term of the sequence. $$ \frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4}, \frac{1}{4}-\frac{1}{5}, \frac{1}{5}-\frac{1}{6}, \ldots $$

Step-by-Step Solution

Answer

The \(n\)th term of the sequence is \( \frac{1}{(n+1)(n+2)} \).

Step 1: Identify the Pattern

Observe the sequence terms: \( \frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4}, \frac{1}{4}-\frac{1}{5}, \ldots \). Notice each term is of the form \( \frac{1}{n} - \frac{1}{n+1} \).

Step 2: Generalize the Pattern

Write the general term formula based on the identified pattern: \( a_n = \frac{1}{n+1} - \frac{1}{n+2} \).

Step 3: Simplify the General Term

Simplify the expression for the \(n\)th term: \[\frac{1}{n+1} - \frac{1}{n+2} = \frac{(n+2) - (n+1)}{(n+1)(n+2)}\] This simplifies to: \[\frac{1}{(n+1)(n+2)}\]

Step 4: Verify the Formula

Check the formula with the first few terms given in the sequence to ensure accuracy. 1. For \( n=1 \): \( \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \) Formula: \( \frac{1}{(1+1)(1+2)} = \frac{1}{6} \)2. For \( n=2 \): \( \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \) Formula: \( \frac{1}{(2+1)(2+2)} = \frac{1}{12} \)The formula matches the sequence.

Key Concepts

nth term

In the context of sequences, the "nth term" refers to a formula that allows us to find any term in the sequence without enumerating all the preceding terms. This is like having a magic spell that directly points to the number we're interested in. Let's break it down with our sequence:

• The sequence in question is: \( \frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4}, \frac{1}{4}-\frac{1}{5}, \ldots \).

The formula for the general term of the sequence, known as the "nth term", is given by:

• \( a_n = \frac{1}{n+1} - \frac{1}{n+2} \).
• Upon simplifying, it becomes \( a_n = \frac{1}{(n+1)(n+2)} \).

This formula enables us to compute any term in the sequence directly, making it simple to predict future terms or verify existing ones.
Finding these formulas often involves spotting patterns, as initially demonstrated by observing the sequence structure. This insight into patterns helps derive a mathematical expression that reflects the sequence's behavior.

telescoping series

A telescoping series is a fascinating concept where successive terms cancel each other out in a collapsing manner. Imagine a telescope extending and collapsing back—this is precisely what happens with our series here. Let's examine how this applies:

• The sequence we're dealing with is \( \frac{1}{n+1} - \frac{1}{n+2} \).

The beauty of a telescoping series lies in its partial sum simplification.
When terms cancel each other out, the remainder for the sum simplifies significantly.

• For a telescoping sequence, most terms will eliminate parts of each other.
• This results in a series where the nth partial sum becomes much easier to handle.

This cancellation is what makes telescoping series so powerful in simplifying complex series problems—leaving frequently only a few terms to manage. It elegantly reduces the challenge of summing multiple fractions to handling just a few, which is a delightful trick of calculus and number theory.

sequences in calculus

Sequences play a critical role in calculus, serving as the building blocks for understanding more complex concepts like limits, continuity, and functions. The sequence that we're dissecting illustrates these ideas beautifully. Let’s dive into why they are important:

• Sequences provide a foundation to introduce limits, one of calculus's core components.
• By examining how a sequence behaves as it approaches infinity, students learn a great deal about limiting behavior.

In our specific scenario, our sequence \( a_n = \frac{1}{(n+1)(n+2)} \) builds a framework for predicting limits:

• As \( n \) becomes very large, \( a_n \) approaches zero, demonstrating convergence.

This concept is crucial in calculus where the behavior of sequences as they grow infinitely large helps in crafting the understanding of infinite series, differentiation, and integration.
Learning about sequences paves a strong foundation for tackling other topics. It blends pattern recognition with mathematical reasoning, which is the essence of mathematical thinking. Understanding sequences, therefore, is fundamental for students aiming to grasp calculus comprehensively.