Problem 12 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{n(n+1)}=\frac{n}{(n+1)}$$

Step-by-Step Solution

Answer

The formula is true for all natural numbers \( n \) by induction.

Step 1: Base Case

First, let's verify the statement for the smallest natural number, which is \( n = 1 \). Substitute \( n = 1 \) into the series: \[ \frac{1}{1 \cdot 2} = \frac{1}{2} \]Now check the right side of the equation when \( n = 1 \):\[ \frac{1}{2} = \frac{1}{2} \]Since both sides are equal, the base case holds.

Step 2: Induction Hypothesis

Assume that the statement is true for some arbitrary positive integer \( k \). That is,\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \]This is the induction hypothesis we will use to prove the next step.

Step 3: Induction Step

We need to show that the statement holds for \( n = k + 1 \). Consider the series:\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} \]Using the induction hypothesis, we have:\[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} \]Combine these fractions:\[ \frac{k(k+2) + 1}{(k+1)(k+2)} \]This simplifies to:\[ \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} \]Which further simplifies to:\[ \frac{k+1}{k+2} \]This matches the right side of the original statement for \( n = k+1 \). Thus, the statement holds for \( n = k+1 \).

Step 4: Conclusion

Since we have proved the base case and shown that if the statement holds for an arbitrary case \( n = k \), it also holds for \( n = k + 1 \), the statement is true for all natural numbers \( n \) by the principle of mathematical induction.

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers beginning from 1 and extending indefinitely: 1, 2, 3, 4, and so on. These numbers are used primarily for counting and ordering. Natural numbers do not include zero, negative numbers, fractions, or decimals. When mathematicians refer to proofs or formulas involving natural numbers, they mean these positive integers.
This concept is essential in many mathematical theorems and proofs, including the one we're discussing here, as it involves proving a statement for all natural numbers. Using this set ensures that our proof applies broadly and follows a logical, well-understood progression of numbers.

Series

A series is the sum of the elements of a sequence of numbers. In math, when we talk about a series, we're often interested in finding the sum of a specific number of terms, or potentially all terms, if the series is infinite. Our problem involves a finite series:

• Each term in the series is expressed as \( \frac{1}{n(n+1)} \).
• The series continues, adding each subsequent term until a given \( n \).

Understanding the basics of series is crucial, as it allows you to see how individual terms contribute to the whole and how they can be manipulated algebraically to simplify expressions, or in this case, to demonstrate an induction hypothesis.

Induction Hypothesis

The induction hypothesis is a crucial assumption within the mathematical induction method. It involves assuming that a given statement is true for a particular natural number, usually denoted as \( k \). For our example, we've assumed:

• \( \frac{1}{1 \cdot 2} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \)

This assumption helps bridge the base case to future cases. If you manage to show that assuming true for \( n = k \) implies truth for \( n = k + 1 \), then by induction, the statement must be true for all natural numbers.
Understanding the role of the induction hypothesis is vital as it provides the foundation upon which the entire inductive proof is built.

Base Case

The base case is the initial step in proving statements using mathematical induction. It involves verifying that the assertion holds true for the first natural number, typically \( n = 1 \). This step establishes the starting point of the proof.
For instance, in our exercise:

• We checked \( \frac{1}{1 \cdot 2} = \frac{1}{2} \).
• The right side was also \( \frac{1}{2} \).
• Both sides were equal, confirming the base case.

Completing a valid base case is essential because it confirms that the proposed statement has at least one instance where it holds true, which initiates the induction process. Without it, you cannot correctly extend the logical proof to all natural numbers.