Question

Prove the following statements with either induction, strong induction or proof by smallest counterexample. Prove that \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^{2}} \leq 2-\frac{1}{n}\) for every \(n \in \mathbb{N}\).

Short answer

The inequality \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^{2}} \leq 2-\frac{1}{n}\) for every n in the natural numbers is proven by mathematical induction.

Step-by-step solution

Base Case

Verify the statement for the base case, n=1. For n=1, the left-hand side of the inequality is \(\frac{1}{1} = 1\). The right-hand side of the inequality is \(2-\frac{1}{1} = 1\). So, the inequality holds for the base case.

Inductive Hypothesis

Assume the statement is true for some arbitrary natural number k. That is, \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{k^{2}} \leq 2-\frac{1}{k}\).

Inductive Step

Prove the statement for the next integer, k+1. Consider \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{k^{2}}+ \frac{1}{(k+1)^{2}}\). By the inductive hypothesis, \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{k^{2}} \leq 2-\frac{1}{k}\), so it then suffices to show that \(\frac{1}{(k+1)^{2}} \leq \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}\). Since \((k+1)^{2} = k^{2} + 2k + 1 > k^{2} + k = k(k+1)\), we have \(\frac{1}{(k+1)^{2}} < \frac{1}{k(k+1)}\). Thus, \(\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{k^{2}}+ \frac{1}{(k+1)^{2}} \leq 2-\frac{1}{k} + \frac{1}{k(k+1)} = 2-\frac{1}{k+1}\), completing the inductive step.

Key concepts

Proof by Induction

Proof by induction is a powerful technique commonly used to establish the truth of infinitely many statements, one for each natural number. It is particularly useful in proving statements about sequences and inequalities, like the one given in our exercise. The process involves two main steps: establishing a base case and then proving an inductive step.

The base case serves as the foundation, confirming the statement is true for the initial term in the sequence. Once the base case is verified, the inductive step assumes the statement is true for some arbitrary natural number and then demonstrates its truth for the next natural number. If these steps are successfully executed, induction allows us to conclude that the statement holds for all natural numbers.

Strong Induction

Although the given exercise uses proof by induction, it's valuable to understand strong induction as another flavor of inductive proofs. Strong induction differs slightly in that the inductive step assumes not just the statement for a single arbitrary case but rather assumes it is true for all cases up to that arbitrary case.

This assumption is often represented as the statement holding for all natural numbers less than or equal to some arbitrary number. Strong induction can be particularly helpful when the inductive step relies on multiple preceding cases, unlike the simple induction method used in our exercise.

Inequality Proof

Inequality proofs are used to show that one expression is always greater than or less than another, within a certain range of values. In our exercise, we aim to prove an inequality involving a sum of a series.

Similar to the way proof by induction progresses, proving an inequality often starts with showing that it holds for a base case, then moving on to demonstrate that if the inequality holds for one case, it will also hold for the next in a sequential order. The intricacies of maintaining the 'less than' or 'greater than' relationship throughout the argument make such proofs a fundamental exercise in mathematical rigor and logic.

Mathematical Proof

A mathematical proof is a logical argument demonstrating the truth of a given proposition or statement beyond any doubt. Constructing a proof is an essential skill in mathematics, involving clear definitions, axioms, propositions, and the logical deduction of one statement from another until the final proposition is shown to be true.

In our case, the proof involves showing an inequality for a series sum, and the mathematical proof requires not just intuition but a careful step-by-step argument that leaves no room for ambiguity or contradiction. Proofs can vary greatly in method and complexity, from direct proofs to more complex ones involving induction, contradiction, or contrapositive arguments.

Sequences and Series

Sequences and series are fundamental concepts in mathematics, especially when discussing proof by induction. A sequence is an ordered list of numbers following a particular rule, while a series is the sum of a sequence of terms.

In our exercise, we are dealing with a series where each term is the reciprocal of the square of a natural number. Familiarity with sequences and series is crucial because they often serve as examples in the context of proofs and can represent complex real-world phenomena in simpler mathematical terms. Understanding their behavior and properties is essential for solving many problems in mathematics and its applications.