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Question Number 144483 by alcohol last updated on 25/Jun/21

$$\left(p_n\right)=(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})$$ $$\underset{k=1}{\sum ^n}{k}^2=\frac{1}{6}n(2n+1)(n+1)$$ $$showthat$$ $$\frac{1}{2}(1+\frac{1}{n})-\frac{1}{12n^2}(2n+1)(n+1)<ln\left(p_n\right)<\frac{1}{2}(1+\frac{1}{n})$$ $$hencefind\underset{n\to \mathrm{\infty }}{\lim }\left(p_n\right)$$ $$2)showthat$$ $$t-\frac{t^2}{2}⩽ln(1+t)⩽t,\mathrm{\forall }t>0$$ $$pleasehelp$$ $$$$

Answered by Olaf_Thorendsen last updated on 26/Jun/21

$$\ln (1+x)=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\frac{x^{n+1}}{n+1}$$ $$\Rightarrow x-\frac{x^2}{2}<\ln (1+x)<x\left(1\right)$$ $$p_n=\underset{k=1}{\prod ^n}(1+\frac{k}{n^2})$$ $$\ln p_n=\underset{k=1}{\sum ^n}\ln (1+\frac{k}{n^2})$$ $$\left(1\right):\frac{k}{n^2}-\frac{k^2}{2n^4}<\ln (1+\frac{k}{n^2})<\frac{k}{n^2}$$ $$\underset{k=1}{\sum ^n}(\frac{k}{n^2}-\frac{k^2}{2n^4})<\ln p_n=\underset{k=1}{\sum ^n}\ln (1+\frac{k}{n^2})<\underset{k=1}{\sum ^n}\frac{k}{n^2}$$ $$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{12n^4}<p_n<\frac{n(n+1)}{2n^2}$$ $$\frac{1}{2}(1+\frac{1}{n})-\frac{(n+1)(2n+1)}{12n^3}<p_n<\frac{1}{2}(1+\frac{1}{n})$$

Commented byalcohol last updated on 26/Jun/21

$$thankyoubrother$$

Commented bypuissant last updated on 07/Jul/21

$$\mathrm{Desole}\mathrm{mais}\mathrm{il}\mathrm{ya}\mathrm{erreur}{\mathrm{c}}^{\prime }\mathrm{est}\ln \left({\mathrm{p}}_{\mathrm{n}}\right)\mathrm{qui}$$ $$\mathrm{est}\mathrm{encadr}\acute{\mathrm{e}}..$$ $$\mathrm{ainsi}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{p}}_{\mathrm{n}}\right)=\frac{1}{2}$$ $$\mathrm{donc}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\left({\mathrm{p}}_{\mathrm{n}}\right)=\sqrt{\mathrm{e}}.$$

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