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Question Number 63919 by raj last updated on 11/Jul/19

$$\underset{n=2}{\prod ^{\mathrm{\infty }}}(1-\frac{1}{n^2})=?$$

Commented by Prithwish sen last updated on 11/Jul/19

$$\underset{\mathrm{n}=2}{\prod ^{\mathrm{n}=\mathbf{n}}}(1-\frac{1}{{\mathrm{n}}^2})=\frac{1.2.3^2.4^2...........\mathrm{n}.(\mathrm{n}+1)}{1^2.2^2.3^2..........{(\mathrm{n}-1)}^2.{\mathrm{n}}^2}$$$$={\left[\frac{(\mathrm{n}+1)!}{\mathrm{n}!}\right]}^2.\frac{1}{2\mathrm{n}(\mathrm{n}+1)}=\frac{{(\mathrm{n}+1)}^2}{2\mathrm{n}(\mathrm{n}+1)}$$$$=\frac{1}{2}(1+\frac{1}{\mathrm{n}})\to \frac{1}{2}\mathrm{as}\mathrm{n}\to \mathrm{\infty }$$$$\mathrm{please}\mathrm{check}.$$

Commented by mathmax by abdo last updated on 11/Jul/19

$$let{A}_n=\prod _{k=2}^n(1-\frac{1}{k^2})\Rightarrow \prod _{k=2}^{\mathrm{\infty }}(1-\frac{1}{k^2})={lim}_{n\to +\mathrm{\infty }}{A}_n$$$$but{A}_n=\prod _{k=2}^n\frac{k^2-1}{k^2}=\prod _{k=2}^n\frac{k-1}{k}\frac{k+1}{k}$$$$=\prod _{k=2}^n\frac{k-1}{k}\prod _{k=2}^n\frac{k+1}{k}=\frac{1}{2}\frac{2}{3}...\frac{n-2}{n-1}\frac{n-1}{n}\times \frac{3}{2}.\frac{4}{3}.....\frac{n}{n-1}\frac{n+1}{n}$$$$=\frac{1}{n}\frac{n+1}{2}=\frac{n+1}{2n}\Rightarrow {lim}_{n\to +\mathrm{\infty }}{A}_n=\frac{1}{2}$$

Commented by raj last updated on 11/Jul/19

$$thankyou$$

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