Problem 35 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. $$\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{n^{2}}\right)=\frac{n+1}{2 n}, \text { if } n \geq 2$$

Step-by-Step Solution

Answer

The statement is true for all \( n \geq 2 \).

Step 1: Base Case

First, verify the statement for the smallest value of \( n \) in the domain, which is \( n = 2 \).Evaluate the left-hand side:\[\left(1 - \frac{1}{2^2}\right) = \left(1 - \frac{1}{4}\right) = \frac{3}{4}\]Evaluate the right-hand side:\[\frac{n+1}{2n} = \frac{2+1}{2 \times 2} = \frac{3}{4}\]Both expressions equal \( \frac{3}{4} \), so the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some \( n = k \), i.e.,\[\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdots\left(1-\frac{1}{k^2}\right) = \frac{k+1}{2k}\]

Step 3: Inductive Step

Prove the statement for \( n = k+1 \), given that it holds for \( n = k \).The left-hand side for \( n = k+1 \) is:\[\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdots\left(1-\frac{1}{k^2}\right)\left(1-\frac{1}{(k+1)^2}\right)\]Using the inductive hypothesis, this can be written as:\[\frac{k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right)\]Simplify the expression:\[\frac{k+1}{2k} \times \left(\frac{(k+1)^2 - 1}{(k+1)^2}\right) = \frac{k+1}{2k} \times \frac{k^2 + 2k}{(k+1)^2}\]Rationalizing, we obtain:\[\frac{(k+1)(k+1)}{2k(k+1)} = \frac{(k+2)}{2(k+1)}\]Thus, \[\left(1 - \frac{1}{4}\right)\cdots\left(1-\frac{1}{(k+1)^2}\right) = \frac{k+2}{2(k+1)}\]This completes the inductive step.

Step 4: Conclusion

Since the base case holds true and the inductive step shows the statement holds for \( n = k+1 \) given it is true for \( n = k \), the statement is proven by induction for all \( n \geq 2 \).

Key Concepts

Base Case

The concept of the base case in mathematical induction involves verifying the truth of a statement for the smallest value of the variable, usually denoted as \( n \). In our original exercise, the domain begins at \( n = 2 \).
This means we must first ensure the formula is valid when \( n = 2 \). By substituting \( n = 2 \) into the expression, we evaluate both sides of the equation.

• The left side: \( \left(1 - \frac{1}{2^2}\right) = \left(1 - \frac{1}{4}\right) = \frac{3}{4} \).
• The right side: \( \frac{n+1}{2n} = \frac{2+1}{2 \times 2} = \frac{3}{4} \).

By comparing both sides, they equate to \( \frac{3}{4} \), so the base case is satisfied. Confirming this initial step is crucial because it acts as the foundation for the entire proof process.

Inductive Hypothesis

The inductive hypothesis is a critical step where we assume that the statement holds true for a particular, yet arbitrary, case \( n = k \). This assumption is not yet proven but is a fundamental part of working forward.
The idea is to believe, without doubt, that the statement:

• \( \left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdots\left(1-\frac{1}{k^2}\right) = \frac{k+1}{2k} \)

holds as true. This means you've pictured it as already valid for this instance.
This step establishes a premise that will be used to prove that the formula works for all values starting from this point onward. It’s like saying, "Assume this works for some magic number, \( k \), and let's see if it works for the next number."

Inductive Step

In the inductive step, we take the assumption from the inductive hypothesis and show that if the statement holds for \( n = k \), then it must also hold for \( n = k+1 \). This is essentially about proving the bridge from one case to the next.
For the expression in our example, the true test lies in evaluating:

• The left side at \( n = k+1 \): \( \left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdots\left(1-\frac{1}{k^2}\right)\left(1-\frac{1}{(k+1)^2}\right) \).
• Using the hypothesis, transform it: \( \frac{k+1}{2k} \times \left(1-\frac{1}{(k+1)^2}\right) \).

Now let's simplify: \( \frac{k+1}{2k} \times \frac{(k+1)^2 - 1}{(k+1)^2} \). After further simplification and rationalization, we find the expression simplifies to \( \frac{k+2}{2(k+1)} \).
Hence, since both sides show equality, our assumption has been successfully extended to \( n = k+1 \), thereby confirming the statement holds true for this successive value. This truly fulfills the essence of induction by paving a way from the assumed \( k \) to the guaranteed \( k+1 \).