Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} \equiv \frac{n}{n+1} $$

Short answer

The statement is true for all natural numbers n by Mathematical Induction.

Step-by-step solution

Understand the Base Case

Verify the statement for the base case, where n=1. For n=1, the left-hand side of the equation is \ \( \frac{1}{1 \cdot 2} = \frac{1}{2} \). The right-hand side is \ \( \frac{1}{1+1} = \frac{1}{2} \). Since both sides are equal, the base case holds.

Assume the Inductive Hypothesis

Assume that the statement is true for some natural number k. That is, \ \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \) holds true.

Prove the Statement for k+1

We need to show that the statement holds for n = k+1, i.e., \ \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} \equiv \frac{k+1}{(k+1)+1} \).

Manipulate the Expression

Using the inductive hypothesis, the left-hand side can be written as: \ \( \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} \). Find a common denominator: \ \( \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} \). Simplifying this gives \ \( \frac{k+1}{k+2} \). Thus, the left-hand side equals the right-hand side, completing the proof for k+1.

Conclusion

Since the base case holds and the induction step is proven, by the Principle of Mathematical Induction, the statement \ \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \equiv \frac{n}{n+1} \) is true for all natural numbers n.

Key concepts

Base Case

Understanding the base case is the first critical step in applying the Principle of Mathematical Induction. The base case verifies that the given statement holds true for the smallest natural number, which is usually 1.
In this exercise, for n = 1, the left-hand side of the equation is \(\frac{1}{1 \cdot 2} \), which simplifies to \(\frac{1}{2}\).
The right-hand side is \(\frac{1}{1 + 1} = \frac{1}{2}\).
Since both sides are equal, the statement holds true for n = 1. This is the foundation that we build upon in the following steps.

Inductive Hypothesis

The inductive hypothesis involves assuming that the statement is true for some arbitrary natural number k.
This assumption is a crucial part of the induction process because it allows us to establish a link between the case we know to be true (the base case) and the next case that we want to prove.
In this exercise, we assume that \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1}\) holds true.
This hypothesis is used in the next step to prove the statement for the next natural number, k+1.

Induction Step

The induction step is the heart of the mathematical induction process.
Here, we prove that if the statement is true for some natural number k, then it is also true for k+1. Using the inductive hypothesis, we need to demonstrate that \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{(k+1)+1}\).
By substituting the inductive hypothesis into the equation, we get \(\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}\).
Finding a common denominator:
\(\frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}\).
This simplifies to \(\frac{k+1}{k+2}\), thus proving the statement for k+1.

Natural Numbers

Natural numbers are the basic building blocks in mathematical induction.
They include all positive integers starting from 1, 2, 3, and so on. Each step in the induction process relies on the properties of natural numbers.
It's essential to understand that for mathematical induction to work, the sequence must follow the set of natural numbers.
For example, starting from the base case at n=1 and moving step-by-step through 2, 3, 4, etc., ensures the proof covers each natural number in the sequence.
This exercise demonstrates the principle of induction applied to a sequence involving natural numbers, ensuring the given statement holds true for each one.