Question

The reciprocal of the value of
${lim}_{n\to \infty }(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})....(1-\frac{1}{n^2})$ is ___________

Answer 1

${lim}_{n\to \infty }\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right).....\left(\frac{n^2-1}{n^2}\right)$

$={lim}_{n\to \infty }(\frac{(2-1)(2+1)}{2\times 2}\times \frac{(3-1)(3+1)}{3\times 3}\times ...\times \frac{(n-1)(n+1)}{n\times n})$

$={lim}_{n\to \infty }\left(\frac{n+1}{2n}\right)={lim}_{n\to \infty }\left(\frac{1+1/n}{2}\right)$

$=\frac{1+0}{2}=\frac{1}{2}$

${\left(\frac{1}{2}\right)}^{-1}=2$