Step 1: Problem Overview:
The task is to find the limit of a product. This product is a telescoping product, meaning intermediate terms will cancel out, simplifying the calculation.
Step 2: Core Technique:
Rewrite the general term \( \left(1-\frac{1}{k^2}\right) \) using the difference of squares: \( a^2 - b^2 = (a-b)(a+b) \).
\[ 1 - \frac{1}{k^2} = \frac{k^2 - 1}{k^2} = \frac{(k-1)(k+1)}{k \cdot k} \]\
Step 3: Detailed Solution:
Define the partial product \(P_n\):
\[ P_n = \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right) ... \left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \left(1-\frac{1}{k^2}\right) \]\
Apply the factorization from Step 2:
\[ P_n = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \prod_{k=2}^{n} \left(\frac{k-1}{k}\right) \left(\frac{k+1}{k}\right) \]\
Expand the product to identify the telescoping pattern:
\[ P_n = \left(\frac{1}{2} \cdot \frac{3}{2}\right) \cdot \left(\frac{2}{3} \cdot \frac{4}{3}\right) \cdot \left(\frac{3}{4} \cdot \frac{5}{4}\right) ... \left(\frac{n-1}{n} \cdot \frac{n+1}{n}\right) \]\
Rearrange terms to group cancelling parts:
\[ P_n = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} ... \frac{n-1}{n}\right) \cdot \left(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} ... \frac{n+1}{n}\right) \]\
The first parenthesis simplifies to \(\frac{1}{n}\).
The second parenthesis simplifies to \(\frac{n+1}{2}\).
Therefore, the partial product is:
\[ P_n = \left(\frac{1}{n}\right) \cdot \left(\frac{n+1}{2}\right) = \frac{n+1}{2n} \]\
Find the limit as \( n \to \infty \):
\[ S = \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{n+1}{2n} \]\
Divide 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: Answer:
The limit \(S\) equals \(\frac{1}{2}\).