Question Number 210044 by maths_plus last updated on 29/Jul/24
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)…\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=??? \\ $$
Commented by Frix last updated on 29/Jul/24
$$\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\:=\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
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
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\centerdot…\centerdot\frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }\centerdot\frac{{n}^{\mathrm{2}} −\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}\centerdot\cancel{\mathrm{3}}}{\mathrm{2}^{\cancel{\mathrm{2}}} }\centerdot\frac{\cancel{\mathrm{2}}\centerdot\cancel{\mathrm{4}}}{\cancel{\mathrm{3}^{\mathrm{2}} }}\centerdot\frac{\cancel{\mathrm{3}}\centerdot\cancel{\mathrm{5}}}{\cancel{\mathrm{4}^{\mathrm{2}} }}\centerdot…\centerdot\frac{\cancel{\left({n}−\mathrm{2}\right){n}}}{\cancel{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }}\centerdot\frac{\cancel{\left({n}−\mathrm{1}\right)}\left({n}+\mathrm{1}\right)}{{n}^{\cancel{\mathrm{2}}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}+\mathrm{1}}{\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$