Problem 16 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\dots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$$

Step-by-Step Solution

Answer

The statement is true for all positive integers using mathematical induction.

Step 1: Base Case

The base case is to check if the statement holds for the smallest positive integer, which is \( n = 1 \). Substitute \( n = 1 \) into the statement:\[\frac{1}{2} = 1 - \frac{1}{2}\]Both sides equal \( \frac{1}{2} \), so the base case is true.

Step 2: Inductive Hypothesis

Assume the statement is true for \( n = k \), that is:\[\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \, \ldots \, + \frac{1}{2^{k}} = 1 - \frac{1}{2^{k}}.\]This is the inductive hypothesis.

Step 3: Inductive Step

We need to show that if the formula holds for \( n = k \), it also holds for \( n = k+1 \). Based on the inductive hypothesis:\[\left( \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k}} \right) + \frac{1}{2^{k+1}} = \left( 1 - \frac{1}{2^{k}} \right) + \frac{1}{2^{k+1}}\]Simplify the right side:\[= 1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k}} + \frac{1}{2^{k}}\cdot\frac{1}{2} = 1 - \frac{1}{2^{k}}\left(1 - \frac{1}{2}\right) = 1 - \frac{1}{2^{k+1}}\]Thus, the statement holds for \( n = k+1 \).

Step 4: Conclusion

By mathematical induction, since the base case is true and the inductive step holds, the statement \( \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}} \) is true for all positive integers \( n \).

Key Concepts

Base Case

In mathematical induction, the base case is the starting point. This involves checking the validity of a statement for the smallest value, usually when \( n = 1 \). This step ensures that the statement is initially true. Let’s look at our given statement: \( \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}} \). For \( n = 1 \), we substitute directly:- Substitute \( n = 1 \) into the statement: \( \frac{1}{2} = 1 - \frac{1}{2} \).Both sides equate to \( \frac{1}{2} \). Therefore, the base case succeeds because the expression holds true at the smallest positive integer \( n \). This confirmation allows us to move forward confidently, knowing our statement has a solid foundation.

Inductive Hypothesis

The inductive hypothesis is the next component in the induction process. This assumption involves taking the statement as correct for \( n = k \), a hypothetical positive integer.In our specific exercise, we propose that the statement holds true for some integer \( k \):- \( \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k}} = 1 - \frac{1}{2^{k}} \).This assumption does not prove the statement outright; rather, it sets the stage for the crucial inductive step. In mathematical circles, making this initial assumption is known as the 'leap of faith'. This step primes us to prove that if the statement works for \( k \), then it will also work for \( k+1 \). We temporarily hold onto this assumption, pressing onward to demonstrate its validity in the next stages.

Inductive Step

The inductive step is where the magic of mathematical induction happens. This phase is about proving that if our statement is accurate for \( n = k \), it must also be true for \( n = k+1 \).Here's how it unfolds for our exercise:- Given the inductive hypothesis: \( \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k}} = 1 - \frac{1}{2^{k}} \)- Demonstrate it works for \( n = k+1 \) by adding \( \frac{1}{2^{k+1}} \) to both sides: \[ \left( 1 - \frac{1}{2^{k}} \right) + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} \]By manipulating the algebraic expression, we confirm the truth of the statement:- \( 1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} \)Thus, logical reasoning follows through that the step from \( k \) to \( k+1 \) is true. Successfully showing this makes the full transition from our base case up to any integer \( n \), signaling that the statement holds universally under these conditions. This step is integral because it stitches our prior assumptions with actual proof, illustrating the perpetual validity of the formal statement.