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Question Number 204270 by universe last updated on 10/Feb/24

Answered by witcher3 last updated on 10/Feb/24

$$\left(1\right)$$$$\mathrm{withe}\mathrm{recursion}\mathrm{We}\mathrm{can}\mathrm{easly}\mathrm{proov}\mathrm{that}{\mathrm{a}}_{\mathrm{n}}>0$$$${\mathrm{a}}_{\mathrm{n}+1}=\mathrm{f}\left({\mathrm{a}}_{\mathrm{n}}\right);\mathrm{f}\left(\mathrm{x}\right)=\ln (1+\mathrm{x})\mathrm{increase}\mathrm{function}$$$${\mathrm{a}}_{\mathrm{n}}>0\Rightarrow {\mathrm{a}}_{\mathrm{n}+1}>\mathrm{f}\left(0\right)=0$$$$\mathrm{f}\left(\mathrm{x}\right)=\ln (1+\mathrm{x});{\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\frac{1}{{(1+\mathrm{x})}^2}<0\mathrm{conncave}\mathrm{function}$$$$\mathrm{tangent}\mathrm{in}\mathrm{zero}\mathrm{y}={\mathrm{f}}^{\prime }\left(0\right)(\mathrm{x}-0)+\mathrm{f}\left(0\right)=\mathrm{x}$$$$\mathrm{we}\mathrm{have}\ln (1+\mathrm{x})<\mathrm{x}\Rightarrow {\mathrm{a}}_{\mathrm{n}+1}=\ln (1+{\mathrm{a}}_{\mathrm{n}})<{\mathrm{a}}_{\mathrm{n}}$$$${\mathrm{a}}_{\mathrm{n}}\mathrm{decrease}\mathrm{and}{\mathrm{a}}_{\mathrm{n}}>0\Rightarrow {\mathrm{a}}_{\mathrm{n}}\mathrm{cv}\mathrm{use}\mathrm{theorem}\mathrm{cv}$$$$\text{Double subscripts: use braces to clarify}$$$$\ln (1+\mathrm{x})=\mathrm{x}\mathrm{use}\mathrm{f}\mathrm{is}\mathrm{concave}\Rightarrow \mathrm{x}=0$$$$\lim {\mathrm{a}}_{\mathrm{n}}=0$$$${\mathrm{a}}_{\mathrm{n}+1}=\ln (1+{\mathrm{a}}_{\mathrm{n}})={\mathrm{a}}_{\mathrm{n}}-\frac{{\mathrm{a}}_{\mathrm{n}}^2}{2}+\mathrm{o}\left({\mathrm{a}}_{\mathrm{n}}^3\right)...\mathrm{Dl}\mathrm{f}\left(\mathrm{x}\right)\mathrm{near}0$$$${\mathrm{a}}_{\mathrm{n}+1}^{-1}-{\mathrm{a}}_{\mathrm{n}}^{-1}={({\mathrm{a}}_{\mathrm{n}}-\frac{{\mathrm{a}}_{\mathrm{n}}^2}{2}+\mathrm{o}\left({\mathrm{a}}_{\mathrm{n}}^3\right))}^{-1}-{\mathrm{a}}_{\mathrm{n}}^{-1}$$$$={\mathrm{a}}_{\mathrm{n}}^{-1}{(1-\frac{{\mathrm{a}}_{\mathrm{n}}}{2}+\mathrm{o}\left({\mathrm{a}}_{\mathrm{n}}^2\right))}^{-1}-{\mathrm{a}}_{\mathrm{n}}^{-1}$$$$={\mathrm{a}}_{\mathrm{n}}^{-1}(1+\frac{{\mathrm{a}}_{\mathrm{n}}}{2}+\mathrm{o}\left({\mathrm{a}}_{\mathrm{n}}^2\right))-{\mathrm{a}}_{\mathrm{n}}^{-1}=\frac{1}{2}+\mathrm{o}\left({\mathrm{a}}_{\mathrm{n}}\right)$$$$\Rightarrow {\mathrm{a}}_{\mathrm{n}+1}^{-1}-{\mathrm{a}}_{\mathrm{n}}^{-1}\sim \frac{1}{2}$$$$\Rightarrow \underset{\mathrm{k}=\mathrm{n}}{\sum ^{\mathrm{N}}}{\mathrm{a}}_{\mathrm{k}+1}^{-1}-{\mathrm{a}}_{\mathrm{k}}^{-1}\sim \mathrm{\Sigma }\frac{1}{2}$$$$\Rightarrow {\mathrm{a}}_{\mathrm{N}+1}^{-1}-{\mathrm{a}}_{\mathrm{n}}^{-1}\sim \frac{\mathrm{N}-\mathrm{n}}{2}\sim \frac{\mathrm{N}}{2}$$$$\Rightarrow {\mathrm{a}}_{\mathrm{N}+1}^{-1}\sim \frac{\mathrm{N}}{2}\sim \frac{\mathrm{N}+1}{2}$$$$\Rightarrow \underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{{\mathrm{a}}_{\mathrm{n}}^{-1}}{\mathrm{n}}=\frac{1}{2}$$$$\text{Double subscripts: use braces to clarify}$$$$$$

Answered by witcher3 last updated on 10/Feb/24

$$\mathrm{for}\left(\mathrm{b}\right)$$$$\frac{1}{\ln (1+\mathrm{x})}-\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}+\mathrm{o}\left({\mathrm{x}}^3\right)}-\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{x}}(\frac{1}{1-(\frac{\mathrm{x}}{2}-\frac{{\mathrm{x}}^2}{3}+\mathrm{o}\left({\mathrm{x}}^2\right))}-1)$$$$=\frac{1}{\mathrm{x}}(1+\frac{\mathrm{x}}{2}-\frac{{\mathrm{x}}^2}{12}+\mathrm{o}\left({\mathrm{x}}^2\right)-1)$$$$=\frac{1}{2}-\frac{\mathrm{x}}{12}+\mathrm{o}\left(\mathrm{x}\right)$$$$\mathrm{let}{\mathrm{V}}_{\mathrm{n}}=\frac{1}{{\mathrm{u}}_{\mathrm{n}}}-\frac{\mathrm{n}}{2}$$$${\mathrm{V}}_{\mathrm{n}+1}-{\mathrm{V}}_{\mathrm{n}}\sim \frac{1}{\ln (1+{\mathrm{u}}_{\mathrm{n}})}-\frac{1}{{\mathrm{u}}_{\mathrm{n}}}-\frac{1}{2}\sim -\frac{{\mathrm{u}}_{\mathrm{n}}}{12}+\mathrm{o}\left({\mathrm{u}}_{\mathrm{n}}\right)$$$${\mathrm{u}}_{\mathrm{n}}\sim \frac{2}{\mathrm{n}}$$$${\mathrm{V}}_{\mathrm{n}+1}-{\mathrm{V}}_{\mathrm{n}}\sim -\frac{1}{6\mathrm{n}}+\mathrm{o}\left(\frac{2}{\mathrm{n}}\right)$$$$\Rightarrow \mathrm{\Sigma }{\mathrm{V}}_{\mathrm{n}+1}-{\mathrm{V}}_{\mathrm{n}}\sim -\frac{1}{6}\mathrm{\Sigma }\frac{1}{\mathrm{n}}\sim -\frac{1}{6}\ln \left(\mathrm{n}\right)$$$${\mathrm{V}}_{\mathrm{n}}\sim -\frac{1}{6}\ln \left(\mathrm{n}\right)$$$$\frac{1}{{\mathrm{u}}_{\mathrm{n}}}-\frac{\mathrm{n}}{2}\sim -\frac{\ln \left(\mathrm{n}\right)}{6}\Rightarrow {\mathrm{u}}_{\mathrm{n}}\sim \frac{1}{\frac{\mathrm{n}}{2}-\frac{\ln \left(\mathrm{n}\right)}{6}}=\frac{2}{\mathrm{n}-\frac{\ln \left(\mathrm{n}\right)}{3}}$$$${\mathrm{nu}}_{\mathrm{n}}-2\sim \frac{2\ln \left(\mathrm{n}\right)}{3(\mathrm{n}-\frac{\ln \left(\mathrm{n}\right)}{3})}$$$$\frac{\mathrm{n}({\mathrm{nu}}_{\mathrm{n}}-2)}{\ln \left(\mathrm{n}\right)}\sim \frac{2\mathrm{nln}\left(\mathrm{n}\right)}{3(\mathrm{n}-\frac{\ln \left(\mathrm{n}\right)}{3})\ln \left(\mathrm{n}\right)}=\frac{2}{3}.\frac{1}{1-\frac{\ln \left(\mathrm{n}\right)}{3}}\to \frac{2}{3}$$

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