lim-x-1-1-2-2-1-1-3-2-1-1-n-2-

Question Number 210044 by maths_plus last updated on 29/Jul/24

\lim_{x \to \infty} (1 - \frac{1}{2^{2}}) (1 - \frac{1}{3^{2}}) \dots (1 - \frac{1}{n^{2}}) = ? ? ?

Commented by Frix last updated on 29/Jul/24

\underset{k = 2}{\prod^{n}} (1 - \frac{1}{k^{2}}) = \frac{n + 1}{2 n} \Rightarrow Answer is \frac{1}{2}

Commented by Frix last updated on 29/Jul/24

!!!

Commented by klipto last updated on 29/Jul/24

Answered by depressiveshrek last updated on 29/Jul/24

\lim_{n \to \infty} \frac{2^{2} - 1}{2^{2}} \cdot \frac{3^{2} - 1}{3^{2}} \cdot \dots \cdot \frac{{(n - 1)}^{2} - 1}{{(n - 1)}^{2}} \cdot \frac{n^{2} - 1}{n^{2}}

= \lim_{n \to \infty} \frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdot \dots \cdot \frac{(n - 2) n}{{(n - 1)}^{2}} \cdot \frac{(n - 1) (n + 1)}{n^{2}}

= \lim_{n \to \infty} \frac{n + 1}{2 n}

= \frac{1}{2}