Problem 15 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$\frac{4}{5}+\frac{4}{5^{2}}+\frac{4}{5^{3}}+\dots+\frac{4}{5^{n}}=1-\frac{1}{5^{n}}$$

Step-by-Step Solution

Answer

Proven by induction, it holds for all positive integers \( n \).

Step 1: Base Case

To prove this statement by induction, let's check the base case, when \( n = 1 \). The left hand side of the equation is \( \frac{4}{5} \), and the right hand side is \( 1 - \frac{1}{5} = \frac{4}{5} \). Both sides are equal, thus the base case holds.

Step 2: Inductive Hypothesis

Assume the induction hypothesis is true for some arbitrary positive integer \( k \), so \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} = 1 - \frac{1}{5^k} \).

Step 3: Inductive Step

We need to prove the statement for \( n = k + 1 \). Consider the left hand side for \( k + 1 \): \[ \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} + \frac{4}{5^{k+1}} \] By the inductive hypothesis, this is equal to: \[ 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} \] To simplify this, let's combine the terms: \[ 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^k} + \frac{4}{5} \cdot \frac{1}{5^k} = 1 - \frac{1}{5^k} \left(1 - \frac{4}{5}\right) \] Simplifying further: \[ 1 - \frac{1}{5^k} \cdot \frac{1}{5} = 1 - \frac{1}{5^{k+1}} \] Hence, the equation \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^{k+1}} \) holds.

Step 4: Conclusion

Since the base case holds and the inductive step has been verified, by the principle of mathematical induction, the statement\( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^n} = 1 - \frac{1}{5^n} \) is true for all positive integers \( n \).

Key Concepts

Base Case

In mathematical induction, the base case is the starting point of the proof. It involves demonstrating that the given statement is true for the initial value of the variable, usually denoted as \( n = 1 \). Here, we need to prove the formula \( \frac{4}{5}+\frac{4}{5^{2}}+\frac{4}{5^{3}}+\dots+\frac{4}{5^{n}}=1-\frac{1}{5^{n}} \) holds for \( n = 1 \).

To do this, let's substitute \( 1 \) for \( n \) in both sides of the equation:

• Left Side: \( \frac{4}{5} \)
• Right Side: \( 1 - \frac{1}{5} = \frac{4}{5} \)

Since both sides are equal, the statement holds true for \( n = 1 \). Verifying the base case is crucial because it serves as a foundation for the induction process. It ensures that the statement is indeed valid at the beginning of the sequence, thus establishing the first "step" of the induction.

Inductive Hypothesis

The inductive hypothesis is the assumption made in the middle of a mathematical induction proof. Here, we assume the statement is true for an arbitrary positive integer \( k \). This means we expect:

• \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} = 1 - \frac{1}{5^k} \)

This assumption is not automatically verified for all \( k \) — it's a strategic assumption we make to help prove the next step, the inductive step.

Making this assumption allows us to use it as a stepping stone. Essentially, by assuming the statement's truth at an arbitrary step \( k \), we can prove it for \( k+1 \), thus establishing the statement holds for every integer following our base case and the current step assumed.

Inductive Step

The inductive step is where the actual grit of mathematical induction comes into play. Here, the aim is to show that if the statement holds for \( n = k \), then it also holds for \( n = k+1 \). Using the inductive hypothesis, we have:

\[ \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} = 1 - \frac{1}{5^k} \]

Now, we need to consider what happens for \( k+1 \):

\[ \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^k} + \frac{4}{5^{k+1}} \]

By substituting from the assumption, this becomes:

\[ 1 - \frac{1}{5^k} + \frac{4}{5^{k+1}} \]

Simplifying the expression by combining terms, one arrives at:

• \( 1 - \frac{1}{5^k}\left( 1 - \frac{4}{5} \right) = 1 - \frac{1}{5^{k+1}} \)

Consequently, the expression \( \frac{4}{5} + \frac{4}{5^2} + \cdots + \frac{4}{5^{k+1}} = 1 - \frac{1}{5^{k+1}} \) holds.

The inductive step confirms that assuming the formula true for \( k \) ensures it’s true for \( k+1 \), reinforcing our proof. Successful completion of this step, when combined with a verified base case, completes the proof by induction.