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Question Number 116915 by mathdave last updated on 07/Oct/20

Answered by Lordose last updated on 07/Oct/20

$$\mathbf{\Omega }=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{a}}^{\mathrm{n}}}{1+{\mathrm{a}}^{\mathrm{n}}}$$$$\mathbf{\Omega }=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{a}}^{\mathrm{n}}\mathrm{lna}}{{\mathrm{a}}^{\mathrm{n}}\mathrm{lna}}$$$$\mathbf{\Omega }=1$$

Answered by mathmax by abdo last updated on 08/Oct/20

$${\mathrm{u}}_{\mathrm{n}}=\frac{{\mathrm{a}}^{\mathrm{n}}}{(1+\mathrm{a})(1+{\mathrm{a}}^2)....(1+{\mathrm{a}}^{\mathrm{n}})}\Rightarrow \ln \left({\mathrm{u}}_{\mathrm{n}}\right)=\mathrm{nln}\left(\mathrm{a}\right)-\ln \left(\prod _{\mathrm{k}=1}^{\mathrm{n}}(1+{\mathrm{a}}^{\mathrm{k}})\right)$$$$=\mathrm{nlna}-\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+{\mathrm{a}}^{\mathrm{k}})\mathrm{we}\mathrm{have}{\ln }^{{}^{\prime }}(1+\mathrm{u})=\frac{1}{1+\mathrm{u}}=1-\mathrm{u}+{\mathrm{u}}^2-..\Rightarrow$$$$\ln (1+\mathrm{u})=\mathrm{u}-\frac{{\mathrm{u}}^2}{2}+...\Rightarrow \mathrm{u}-\frac{{\mathrm{u}}^2}{2}⩽\ln (1+\mathrm{u})⩽\mathrm{u}\Rightarrow$$$${\mathrm{a}}^{\mathrm{k}}-\frac{{\mathrm{a}}^{2\mathrm{k}}}{2}⩽\ln (1+{\mathrm{a}}^{\mathrm{k}})⩽{\mathrm{a}}^{\mathrm{k}}\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{a}}^{\mathrm{k}}-\frac{1}{2}{\mathrm{a}}^{2\mathrm{k}})⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+{\mathrm{a}}^{\mathrm{k}})⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{a}}^{\mathrm{k}}\Rightarrow$$$$\Rightarrow \frac{1-{\mathrm{a}}^{\mathrm{n}+1}}{1-\mathrm{a}}-1-\frac{1}{2}\times (\frac{1-{\mathrm{a}}^{2\mathrm{n}+2}}{1-{\mathrm{a}}^2}-1)⩽\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+{\mathrm{a}}^{\mathrm{k}})⩽\frac{1-{\mathrm{a}}^{\mathrm{n}+1}}{1-\mathrm{a}}-1(\mathrm{if}\mathrm{a}\ne 1)$$$$\mathrm{nlna}-\frac{1-{\mathrm{a}}^{\mathrm{n}+1}}{1-\mathrm{a}}+1⩽\ln \left({\mathrm{u}}_{\mathrm{n}}\right)⩽\mathrm{nln}\left(\mathrm{a}\right)-\frac{1-{\mathrm{a}}^{\mathrm{n}+1}}{1-\mathrm{a}}+\frac{1}{2}+\frac{1-{\mathrm{a}}^{2\mathrm{n}+2}}{2(1-{\mathrm{a}}^2)}$$$$\mathrm{if}\mid \mathrm{a}\mid <1\mathrm{due}\mathrm{to}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\mathrm{nlna}=+\mathrm{\infty }\mathrm{we}\mathrm{get}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{u}}_{\mathrm{n}}\right)=+\mathrm{\infty }\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}=+\mathrm{\infty }\mathrm{if}\mid \mathrm{a}\mid >1\mathrm{we}\mathrm{put}\mathrm{a}=\frac{1}{\mathrm{u}}\Rightarrow \mid \mathrm{u}\mid <1\Rightarrow$$$$-\mathrm{nlnu}-\frac{1-\frac{1}{{\mathrm{u}}^{\mathrm{n}+1}}}{1-\frac{1}{\mathrm{u}}}+1⩽\ln \left({\mathrm{u}}_{\mathrm{n}}\right)⩽-\mathrm{nlnu}-\frac{1-\frac{1}{{\mathrm{u}}^{\mathrm{n}+1}}}{1-\frac{1}{\mathrm{u}}}+\frac{1-\frac{1}{{\mathrm{u}}^{2\mathrm{n}+2}}}{2(1-\frac{1}{{\mathrm{u}}^2})}$$$$....\mathrm{be}\mathrm{continued}...$$

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