Question

If \( S = \lim_{n \to \infty} \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right) ... \left(1-\frac{1}{n^2}\right) \), then S is equal to:

• A. 0
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{2}\)
• D. 1

Hint:

When faced with an infinite product, always try to simplify the general term. Factoring using common algebraic identities like the difference of squares is a very common technique that often leads to a telescoping series or product.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves finding the limit of a product of terms. This type of product is often a "telescoping product," where intermediate terms cancel out, simplifying the expression significantly.

Step 2: Key Formula or Approach:
The key is to rewrite the general term of the product, \( \left(1-\frac{1}{k^2}\right) \), using the difference of squares formula: \( a^2 - b^2 = (a-b)(a+b) \).
\[ 1 - \frac{1}{k^2} = \frac{k^2 - 1}{k^2} = \frac{(k-1)(k+1)}{k \cdot k} \]
Step 3: Detailed Explanation:
Let \(P_n\) be the partial product: \[ P_n = \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right) ... \left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \left(1-\frac{1}{k^2}\right) \] Using the factorization from Step 2: \[ P_n = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \prod_{k=2}^{n} \left(\frac{k-1}{k}\right) \left(\frac{k+1}{k}\right) \] Let's expand this product to see the telescoping pattern: \[ P_n = \left(\frac{1}{2} \cdot \frac{3}{2}\right) \cdot \left(\frac{2}{3} \cdot \frac{4}{3}\right) \cdot \left(\frac{3}{4} \cdot \frac{5}{4}\right) ... \left(\frac{n-1}{n} \cdot \frac{n+1}{n}\right) \] We can rearrange the terms to group the canceling parts together: \[ P_n = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} ... \frac{n-1}{n}\right) \cdot \left(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} ... \frac{n+1}{n}\right) \] In the first parenthesis, terms cancel out, leaving \(\frac{1}{n}\).
In the second parenthesis, terms cancel out, leaving \(\frac{n+1}{2}\).
So, the partial product simplifies to: \[ P_n = \left(\frac{1}{n}\right) \cdot \left(\frac{n+1}{2}\right) = \frac{n+1}{2n} \] Now, we need to find the limit as \( n \to \infty \): \[ S = \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{n+1}{2n} \] Divide the numerator and denominator by \(n\): \[ S = \lim_{n \to \infty} \frac{1+\frac{1}{n}}{2} = \frac{1+0}{2} = \frac{1}{2} \]
Step 4: Final Answer:
The value of the limit \(S\) is \(\frac{1}{2}\).