Question

Define \[ S = \lim_{n \to \infty} \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\cdots\left(1 - \frac{1}{n^2}\right). \] Then

• A. \(S = \frac{1}{2}\).
• B. \(S = \frac{1}{4}\).
• C. \(S = \frac{1}{3}\).
• D. \(S = \frac{3}{4}\).

Hint:

Infinite product identities like \(\frac{\sin(\pi x)}{\pi x}\) are powerful tools to evaluate products involving \((1 - \frac{x^2}{n^2})\).

Answer 1

The Correct Option is A

Step 1: Express the general term.
We can use the known infinite product identity \[ \frac{\sin(\pi x)}{\pi x} = \prod_{n=1}^{\infty} \left(1 - \frac{x^2}{n^2}\right). \] Setting \(x=1\), we get \[ 0 = \frac{\sin(\pi)}{\pi} = \prod_{n=1}^{\infty}\left(1 - \frac{1}{n^2}\right). \] However, the product starting from \(n=2\) (as given in the question) is \[ \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = \frac{1}{2}. \]
Step 2: Conclusion.
Thus, the required limit \(S = \frac{1}{2}\).