Question

This problem explores ways to prove that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\). a. First, we explore how to prove the formula by contradiction. In other words, assume that there is some integer \(n\) that makes the formula false. In this case, there must be some smallest \(n\) that makes the formula false. i. Can this smallest \(n\) be \(1 ?\) ii. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{i}} $$ when \(i\) is a positive integer smaller than this smallest \(n\) ? iii. Is \(n-1\) a positive integer for this smallest \(n\) ? iv. What do you know about $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n-1}} $$ for this smallest \(n\) ? v. Write the answer to part iv as an equation, add \(2 / 3^{n}\) to both sides, and simplify the right side. vi. What does the equation that results from part \(\mathrm{v}\) say about your assumption that the formula is false? vii. What can you conclude about the truth of the formula? viii. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of deriving your contradiction? b. i. What is the base case in a proof by mathematical induction that $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{n}}=1-\left(\frac{1}{3}\right)^{n} $$ for all positive integers \(n\) ? ii. What would you assume as an inductive hypothesis? iii. What would you prove in the inductive step of a proof of this formula by induction? iv. Prove it. v. What does the principle of mathematical induction allow you to conclude? vi. If \(p(k)\) is the statement $$ \frac{2}{3}+\frac{2}{9}+\cdots+\frac{2}{3^{k}}=1-\left(\frac{1}{3}\right)^{k} $$ what implication did you prove in the process of doing your proof by induction?

Short answer

The formula holds true for all positive integers \( n \) by both contradiction and mathematical induction methods.

Step-by-step solution

Initialize Assumption for Contradiction

Assume that there exists a positive integer \( n \) for which the formula \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^n} = 1 - \left(\frac{1}{3}\right)^n \) is false. Among these integers, let \( n \) be the smallest such integer.

Testing Smallest Integer Assumption

i. Check if \( n = 1 \) can be the smallest integer making the formula false. Calculate both sides for \( n = 1 \). Left side: \( \frac{2}{3} \); Right side: \( 1 - \frac{1}{3} = \frac{2}{3} \). Both sides are equal, so \( n eq 1 \).

Exploring Smaller Integers

ii. For any integer \( i < n \), the statement \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^i} = 1 - \left(\frac{1}{3}\right)^i \) must hold true since \( n \) is assumed to be the smallest integer that is false.

Definition of \( n-1 \)

iii. Since \( n \) is the smallest integer making the statement false, \( n-1 \) must be a positive integer.

Analyzing \( n-1 \)

iv. By the assumption of smallest \( n \), the equation \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^{n-1}} = 1 - \left(\frac{1}{3}\right)^{n-1} \) holds true for \( n-1 \).

Extending Equation for Contradiction

v. Adding \( \frac{2}{3^n} \) to both sides gives \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^{n-1}} + \frac{2}{3^n} = 1 - \left(\frac{1}{3}\right)^{n-1} + \frac{2}{3^n} \). Simplifying the right side confirms the original statement for \( n \), hence contradicting the assumption that \( n \) is the first false integer.

Conclusion on Contradiction

vi. The contradiction implies that no smallest \( n \) exists for which the formula is false, confirming the truth of the formula for all \( n \).

Proving Implication with Contradiction

viii. From the contradiction, it is proven that \( p(k) \) implies \( p(k+1) \) for all positive integers \( k \).

Base Case for Induction

b.i. The base case for induction is \( n = 1 \), where the formula simplifies to \( \frac{2}{3} = 1 - \frac{1}{3} \), which is true.

Inductive Hypothesis

b.ii. Assume the formula holds for some \( k \): \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^k} = 1 - \left(\frac{1}{3}\right)^k \).

Inductive Step

b.iii. We must show the formula holds for \( k+1 \), i.e., \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^k} + \frac{2}{3^{k+1}} = 1 - \left(\frac{1}{3}\right)^{k+1} \).

Proving Inductive Step

b.iv. Using the inductive hypothesis, \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^k} = 1 - \left(\frac{1}{3}\right)^k \). Adding \( \frac{2}{3^{k+1}} \) gives \( 1 - \left(\frac{1}{3}\right)^k + \frac{2}{3^{k+1}} = 1 - \left(\frac{1}{3}\right)^{k+1} \), proving the step.

Conclusion via Induction

b.v. By the principle of mathematical induction, the formula is true for all positive integers \( n \).

Implication from Induction

b.vi. The induction process proves that if the formula holds for \( n = k \), then it holds for \( n = k+1 \).

Key concepts

Proof by Contradiction

Proof by contradiction is a classic mathematical method used to prove that a statement is true by assuming the opposite and then showing that this assumption leads to a contradiction. In the context of the given problem, we assume that there is a smallest positive integer \( n \) for which the formula \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^n} = 1 - \left(\frac{1}{3}\right)^n \) is false. By calculating for \( n = 1 \), we find that both sides of the equation equal \( \frac{2}{3} \), thus \( n eq 1 \).
If \( n \) were the smallest value making the formula false, then for any integer \( i < n \), the formula must still hold, leading to a contradiction. Since assuming \( n \) as the smallest false integer contradicts the proven calculations, we conclude that no such \( n \) exists. Therefore, the formula must be true for all positive integers \( n \).

• Start by assuming the opposite of what you want to prove.
• Find a logical inconsistency in the assumption.
• Use this contradiction to conclude the original statement is true.

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the problem, the series \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^n} \) is geometric, with a first term of \( \frac{2}{3} \) and a common ratio of \( \frac{1}{3} \).
For any geometric series, the sum can be calculated using the formula:
\[ S_n = a \frac{1-r^n}{1-r} \]
Where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Applying this to our series gives:
\[ S(n) = \frac{2}{3} \frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}} = 1 - \left(\frac{1}{3}\right)^n \]
Thus confirming the series sum closely aligns with the formula given, supporting our attempts to verify or prove it.

• Understand the sequence and common ratio.
• Use the geometric series sum formula to find the total.
• Recognize its usefulness in verifying algebraic identities.

Inductive Hypothesis

The inductive hypothesis is an essential part of mathematical induction proofs. It involves assuming that a statement is true for some arbitrary positive integer \( k \). In the context of the problem, we assume that the formula \( \frac{2}{3} + \frac{2}{9} + \cdots + \frac{2}{3^k} = 1 - \left(\frac{1}{3}\right)^k \) holds for \( k \).
With this hypothesis, the goal is to prove that if the formula holds for \( k \), it must also be true for \( k+1 \). This step involves using the assumption for \( k \) and showing it extends logically and mathematically to \( k+1 \). By doing so, the truth of the statement is established for all positive integers.

• Assume the formula or statement holds for a particular \( k \).
• This assumption should be used in the inductive step.
• If successful, it justifies the statement for all subsequent integers.

Base Case

The base case in mathematical induction is the initial step where we prove that the given statement holds true for the smallest possible value, commonly \( n = 1 \). For the exercise in question, the base case involves verifying the formula: \( \frac{2}{3} = 1 - \frac{1}{3} \).
This calculation is straightforward and confirms the formula's truth for the base value.
The reasoning behind verifying the base case is to anchor the induction process in reality. Once the base is shown to be valid, the inductive step, dependent on the assumption that the statement holds for \( n = k \), can take over to prove the statement for all subsequent integers.

• A necessary starting point in induction proofs.
• Validates the first rung in the 'induction ladder.'
• Ensures the logical chain from base to generalized proof remains unbroken.