Question

Solve

$\underset{x\to \infty }{lim}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\right]=$

Answer 1

given that,

$\underset{x\to \infty }{lim}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\right]$

$=\prod _{n=2}^{\infty }\frac{n^2-1}{n^2}$

$=\prod _{n=2}^{\infty }\frac{(n-1)}{n}\frac{(n+1)}{n}$

$=\underset{n\to \infty }{lim}(\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}.......\frac{n+1}{n})$

$=\underset{x\to \infty }{lim}\frac{n+1}{2n}$

$=\underset{n\to \infty }{lim}\frac{1+1/n}{2}=\frac{1}{2}$