Problem 23 - College Algebra

Question

In Exercises \(11-30,\) use mathematical induction to prove that each statement is true for every positive integer \(n\) $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{n(n+1)}=\frac{n}{n+1} $$

Step-by-Step Solution

Answer

By mathematical induction, the given formula is verified to hold true for all positive integers.

Step 1: Base Case

First, verify the base case (i.e., \( n = 1 \)). For \( n = 1 \), the given formula yields \( \frac{1}{1 \cdot 2} = \frac{1}{2} \), which is equal to \( \frac{1}{1 + 1} = \frac{1}{2} \) calculated from the right side. Hence, the statement is true for \( n = 1 \).

Step 2: Inductive Step

Assume that the statement is true for \( n = k \) for some fixed positive integer \( k \). That is, \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{k(k+1)} = \frac{k}{k+1} \] is assumed to be true.

Step 3: Verification Step

To complete the induction, show that the statement holds for \( n = k+1 \) under the assumption that it is true for \( n = k \). The left-hand side for \( n = k + 1 \) is \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} \] Based on the inductive step, the sum of the first \( k \) terms can be replaced with \( \frac{k}{k+1} \), leading to: \[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} \] Simplification of the above expression will lead to \( \frac{k+1}{k+2} \), which matches with the right side of the equation for \( n = k+1 \). Hence, by mathematical induction, the formula is proved to hold for all positive integers.

Key Concepts

Sequence Series

A sequence series, in this context, is a sum of terms defined in a sequential manner. Parts of this series can often be seen in mathematics where terms follow a specific pattern or rule. In the given exercise, each term is represented as \( \frac{1}{n(n+1)} \), where \( n \) is a positive integer that increases as each term is added to the sum. This creates a sequence which is then summed.

Understanding sequence series is crucial, as they form the basis for many mathematical concepts and solutions. These sequences are often infinite or finite, but in this exercise, we focus on a finite number of terms. Visualizing terms as items in a chain can help: each term is linked to its predecessor and successor by the "rule" established for the series.

For example, the first few terms are:

• \( \frac{1}{1 \cdot 2} \)
• \( \frac{1}{2 \cdot 3} \)
• \( \frac{1}{3 \cdot 4} \)

Such series often require methods like proof by induction to prove the formula for any positive integer, as shown in this problem.

Proof by Induction

Proof by induction is a popular mathematical technique often employed to prove statements are true for all positive integers. It works similarly to dominoes lined up in a row.

Here's how:

• Base Case: Start by proving the property for the initial case, usually \( n = 1 \). If this step holds true, it's like knocking over the first domino.
• Inductive Step: Assume the statement is true for \( n = k \), a specific yet arbitrary step.
• Verification Step: Then prove the statement must also be true when stepping from \( n = k \) to \( n = k + 1 \). If this linkage is established, \( n = k + 1 \) holds true as well, leading to an infinite chain reaction proving the statement for all successive integers.

Mathematical induction combines logical reasoning with the assumption technique, ensuring that truths are carried forward incrementally. In the exercise, proving for \( n = 1 \) establishes a sound base and connecting \( k \) to \( k + 1 \) completes this inductive argument.

Positive Integers

In mathematics, positive integers are the set of numbers \( 1, 2, 3, ... \). They are numbers without fractions, decimals, or negatives, and represent quantities like counts, rankings, and sequences.

Positive integers are foundational in various mathematical contexts, especially when proving general statements such as the existence and properties of a sequence series.

For instance, in the exercise, we use positive integers as our domain because each term and step is constructed based on integer values. They help in creating fixed intervals or stages in problems used in proof by induction or sequence analysis. This reliance on positive integers ensures consistent, straightforward applications of formulas and rules in mathematics.

This problem specifically shows how every step of the inductive proof relies on sequential positive integers,

Sum of Series

The sum of a series is the result of adding up all the terms within a sequence. It's an important concept as it lets us calculate the total value that multiple terms collectively hold.

In the given example, the series is:

• \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)} \)

The task is to find its sum, which is posited as \( \frac{n}{n+1} \) for each positive integer \( n \).

Calculating the sum of series involves applying the established formulae to verify its correctness. Proof by induction demonstrates this relationship is valid for any positive integer sequence, as derived in the solution.

The formula for the sum helps in simplifying complex arithmetic into manageable segments, ensuring that we can predict the sum without tedious computation. By understanding how each term contributes to the overall total, mathematicians can use this sum to solve larger, more complex problems efficiently.