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 holds true for all natural numbers \( n \) by mathematical induction.

Step 1: Base Case

First, verify that the formula holds true for the initial value of \( n = 1 \). Substitute \( n = 1 \) into both sides of the formula. The left-hand side is \( \frac{1}{1 \cdot 2} = \frac{1}{2} \) and the right-hand side is \( \frac{1}{1+1} = \frac{1}{2} \). Both sides are equal, so the base case holds true.

Step 2: Inductive Hypothesis

Assume the formula is true for some arbitrary positive integer \( k \). That is assume \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} = \frac{k}{k+1} \) is true.

Step 3: Inductive Step

We now prove that if the formula holds for \( n = k \), it also holds for \( n = k+1 \). Consider the expression for \( n = k+1 \):\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}. \]Using our inductive hypothesis, replace the sum up to \( k \) with \( \frac{k}{k+1} \), so we have:\[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}. \]Combining these fractions, we get:\[ \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}. \]Simplifying gives:\[ \frac{k+1}{k+2}. \]

Step 4: Conclusion

Since we have shown:1. The base case holds true when \( n = 1 \);2. If the formula holds for \( n = k \), it holds for \( n = k+1 \).By the principle of mathematical induction, the formula \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1} \) is true for all natural numbers \( n \).

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers starting from 1 and moving upwards: 1, 2, 3, 4, 5, and so on. These are the numbers we first learn about in mathematics. They are used for counting and ordering. In the context of mathematical induction, natural numbers are crucial because induction requires proving a statement for these numbers.

• Natural numbers do not include zero or any negative numbers.
• They are infinite, meaning the sequence never ends.
• They follow simple arithmetic operations—adding, subtracting, multiplying, without introducing fractions or decimals.
• When using mathematical induction, the statements we prove typically pertain to all natural numbers.

Understanding natural numbers is essential because they serve as the foundation for proving mathematical statements using induction.

Inductive Hypothesis

The inductive hypothesis is a key component in mathematical induction. It involves assuming a given statement is true for a specific but arbitrary natural number, usually denoted as "k." This assumption is what helps bridge the gap to prove the statement for the next natural number.Here's how the inductive hypothesis works:

• You assume the statement holds for an arbitrary positive integer, \( k \).
• This assumption is not a conclusion; it's a temporary step to help prove the next case.
• The hypothesis allows us to apply known true elements to the next stage, making the proof process manageable and strategic.
• Effectively, it's like setting a stepping stone in place so that you can reach the next part of your logical journey.

By formulating a strong inductive hypothesis, you set a path towards proving the truth for the next natural number, \( k + 1 \). It is a critical middle step that must be properly established for mathematical induction to work smoothly.

Base Case

The base case in mathematical induction serves as the starting point of the proof. It is essential to establish that the statement we want to prove is true for an initial value, often \( n = 1 \), before moving forward with the induction steps.The significance of the base case includes:

• It confirms that the formula or statement works when \( n \) takes its smallest possible value in the set of natural numbers.
• This step serves as the grounding truth upon which the rest of the induction is built.
• Without a valid base case, you cannot assume any subsequent steps are valid, making the whole process of induction incomplete.

For example, in our problem, we verify the base case by plugging \( n = 1 \) into the equation to check if both sides match. It's like proving the ground floor of a building is stable before adding more floors.

Inductive Step

The inductive step is where the magic of mathematical induction truly comes into play. This is where you use the inductive hypothesis to show that if the formula is true for \( n = k \), it must also be true for \( n = k+1 \), thereby setting up a chain reaction that proves it's true for all natural numbers.Key elements of the inductive step:

• It involves taking the assumption (inductive hypothesis) and extending it one step further to \( k+1 \).
• This step often requires algebraic manipulation to show how the truth of the statement for \( k \) logically leads to its truth for \( k + 1 \).
• This process demonstrates the domino effect—if one statement is true, the next one logically follows.
• It's critical to carry out these computations accurately, as any mistake might disrupt the entire chain.

To tie it back to the exercise, once the sum is assumed true for \( n = k \), we prove that adding the term \( \frac{1}{(k+1)(k+2)} \) leads to the formula being true at \( n = k+1 \). Successfully handling the inductive step ensures that the formula or statement holds for infinitely many cases.