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Question Number 104174 by Dwaipayan Shikari last updated on 19/Jul/20

$$\underset{\mathrm{n}=1}{\prod ^{\mathrm{\infty }}}{\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)}^2$$

Answered by OlafThorendsen last updated on 19/Jul/20

$${\mathrm{S}}_n=\underset{k=1}{\prod ^n}{\left(\frac{k}{k+1}\right)}^2$$$${\mathrm{S}}_n={\left(\frac{1}{2}\right)}^2{\left(\frac{2}{3}\right)}^2{\left(\frac{3}{4}\right)}^2...{\left(\frac{n}{n+1}\right)}^2$$$${\mathrm{S}}_n=\frac{1}{{(n+1)}^2}$$$$\text{Double subscripts: use braces to clarify}$$

Answered by mathmax by abdo last updated on 19/Jul/20

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}\frac{{\mathrm{k}}^2}{{(\mathrm{k}+1)}^2}\Rightarrow \ln \left({\mathrm{S}}_{\mathrm{n}}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln \left(\frac{{\mathrm{k}}^2}{{(\mathrm{k}+1)}^2}\right)=2\sum _{\mathrm{k}=1}^{\mathrm{n}}\{\ln \left(\mathrm{k}\right)-\ln (\mathrm{k}+1)\}$$$$=2\{\ln 1-\ln \left(2\right)+\ln \left(2\right)-\ln \left(3\right)+...+\ln \left(\mathrm{n}\right)-\ln (\mathrm{n}+1)\}$$$$=-2\ln (\mathrm{n}+1)\to -\mathrm{\infty }\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{S}}_{\mathrm{n}}\right)=0\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=0$$

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