Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. If \(n \in \mathbb{N},\) then \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{n(n+1)}=1-\frac{1}{n+1} .\)

Short answer

Given statement has been proven to be true using mathematical induction. The base case was verified and following the induction hypothesis and inductive step, we have shown that the statement is indeed true for all positive integers.

Step-by-step solution

Base Case Verification

Firstly, let's check if the statement holds true for \(n = 1\). So, substituting \(n = 1\) in the given equation, we get \(1/2\). Also, on the right side of the equation, substituting \(n = 1\) yields \(1 - 1/2\) which also equals \(1/2\). Since both sides of the equation are equal, the statement holds true for \(n = 1\). Hence, the base case is verified.

Induction Hypothesis

Now, let's make an assumption that the statement holds true for \(n = k\), such that \(k \geq 1\). So, we can write this as \(1/(1*2) + 1/(2*3) + ... + 1/(k*(k+1)) = 1 - 1/(k+1)\). We assume this statement to be true. This is our induction hypothesis.

Inductive Step

Next, we need to prove that the statement is true for \(n = k + 1\) assuming it is true for \(n = k\). So, we can write our expression as \(1/(1*2) + 1/(2*3) + ... + 1/(k*(k+1)) + 1/((k+1)*(k+2))\). According to our induction hypothesis, the left part aside from the last term is \(1 - 1/(k+1)\) and so our expression becomes \((1 - 1/(k+1)) + 1/((k+1)*(k+2))\). Simplifying, we get \(1 - 1/(k+2)\) which is the original equation with \(n = k+1\). Hence, we have proven the inductive step.

Key concepts

Induction

Induction is a mathematical proof technique that demonstrates the truth of an infinite sequence of statements. It works in two steps: the base case and the inductive step. First, you show that the statement is true for a starting value, which is the base case. In the exercise, the base case is checked for n=1, ensuring that the statement works at the beginning. This is crucial because if the first domino doesn't fall, the rest won't either.

For the subsequent inductive step, you assume the statement works for an arbitrary natural number k, then show that based on this assumption, the statement also works for k+1. It's like showing if one domino falls, it will knock down the next one. Thus, if the base case holds (the first domino falls), and the inductive step works (each domino knocks down the next), the statement is true for all natural numbers.

Strong Induction

Strong induction slightly tweaks the standard induction's inductive step. Instead of assuming the statement is true for a single arbitrary natural number k, you assume it's true for all natural numbers up to k. Then, you show the statement holds for k+1. This provides a stronger basis for the inductive step because you're not just assuming the previous 'domino' falls; you're ensuring all prior dominos do too, which might be necessary for some proofs where the next step depends on several prior steps, not just the immediate last one.

Proof by Smallest Counterexample

Proof by smallest counterexample is a technique used to prove a statement is true by initially assuming there is a smallest counterexample that would make the statement false and then showing that such a counterexample cannot exist. It is a form of indirect argument, like proof by contradiction, where you show that if the statement were false, it would lead to an absurdity or contradiction, and therefore, the statement must be true. It's like saying if not even a single domino refuses to fall, then none of them can—the entire sequence must proceed as claimed.

Series Summation

Series summation refers to the process of adding up the terms of a series—a list of numbers that often follow a specific pattern. In our exercise, the series is the sum of fractions in the form 1/(n(n+1)). Finding the sum of a series can sometimes involve clever tricks or formulas. In this case, the given series is telescoping, meaning most terms cancel each other out when summed, eventually simplifying to a form that is much easier to evaluate. Summation techniques are crucial in determining exact values of series or establishing properties of the series as a whole.

Mathematical Induction

Understanding Mathematical Induction

Mathematical Induction is a broader term that encompasses the concept of induction and strong induction. Essentially, it's a method of mathematical proof used to establish that a given statement is true for all natural numbers. It mirrors the principle of falling dominos. Once you’ve proven the base case (the first domino falls), and that if the nth statement implies the n+1st statement (every domino causes the next to fall), you’ve effectively shown that all dominos fall, or in other words, the statement is true for all natural numbers due to the chain reaction that starts with the base case progression.

• The exercise provided illustrates mathematical induction, which starts by verifying the formula for n=1.
• Next, assuming the formula is valid for an arbitrary n=k (induction hypothesis).
• The final step shows that the formula is then valid for n=k+1, hence completing the proof by induction.

The concept is one of the foundations of mathematics, particularly in areas like number theory, computer science, and logic.