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Question Number 194960 by Erico last updated on 20/Jul/23

$$\mathrm{Soit}x>1.\mathrm{On}\mathrm{d}\acute{\mathrm{e}finie}\mathrm{la}\mathrm{suite}\left({\mathrm{p}}_{\mathrm{n}}\right)\mathrm{par}$$$${\mathrm{p}}_1=x\mathrm{et}\mathrm{\forall }\mathrm{n}\in {\mathrm{IN}}^{\ast }{\mathrm{p}}_{\mathrm{n}+1}={2\mathrm{p}}_{\mathrm{n}}^2-1$$$$\mathrm{Montrer}\mathrm{que}\underset{\mathrm{n}\to +\mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}(1+\frac{1}{{\mathrm{p}}_{\mathrm{k}}})=\sqrt{\frac{x+1}{x-1}}$$

Answered by witcher3 last updated on 22/Jul/23

$${\mathrm{p}}_{\mathrm{n}}=\mathrm{ch}\left({\mathrm{w}}_{\mathrm{n}}\right);{\mathrm{w}}_{\mathrm{n}}⩾0$$$$\mathrm{x}\stackrel{\mathrm{f}}{\to }\mathrm{ch}\left(\mathrm{x}\right)$$$$\mathrm{bijection}[0,\mathrm{\infty }[\stackrel{\mathrm{f}}{\to }[1,\mathrm{\infty }[$$$$\Rightarrow {\mathrm{p}}_{\mathrm{n}+1}=\mathrm{ch}\left({\mathrm{w}}_{\mathrm{n}+1}\right)$$$${2\mathrm{ch}}^2\left({\mathrm{w}}_{\mathrm{n}}\right)-1=\mathrm{ch}\left({2\mathrm{w}}_{\mathrm{n}}\right)$$$$\mathrm{ch}\left({\mathrm{w}}_{\mathrm{n}+1}\right)=\mathrm{ch}\left({2\mathrm{w}}_{\mathrm{n}}\right)\Rightarrow {\mathrm{w}}_{\mathrm{n}+1}={2\mathrm{w}}_{\mathrm{n}}\mathrm{f}\mathrm{is}\mathrm{injective}$$$${\mathrm{w}}_{\mathrm{n}}=2^{\mathrm{n}-1}{\mathrm{w}}_1,{\mathrm{w}}_1=\mathrm{argch}\left(\mathrm{x}\right)$$$${\mathrm{p}}_{\mathrm{n}}=\mathrm{ch}\left({\mathrm{w}}_1{2}^{\mathrm{n}-1}\right)$$$$\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}(1+\frac{1}{{\mathrm{p}}_{\mathrm{n}}})=\underset{\mathrm{k}=1}{\prod ^{\mathrm{n}}}\left(\frac{1+\mathrm{ch}\left({\mathrm{w}}_1{2}^{\mathrm{n}-1}\right)}{\mathbf{ch}\left({\mathbf{w}}_1{2}^{\mathbf{n}-1}\right)}\right)=\mathrm{\Pi }$$$$(1+{\mathrm{p}}_1).\underset{\mathrm{k}=2}{\prod ^{\mathrm{n}}}\left(\frac{{2\mathrm{ch}}^2\left(2^{\mathrm{k}-2}{\mathrm{w}}_1\right)}{\mathrm{ch}\left({\mathrm{w}}_1{2}^{\mathrm{k}-1}\right)}\right)=\frac{1+{\mathrm{p}}_1}{\mathrm{ch}\left({\mathrm{w}}_1\right)}.2^{\mathrm{n}-1}[.\underset{\mathrm{k}=2}{\prod ^{\mathrm{n}}}\mathrm{ch}\left(2^{\mathrm{k}-2}{\mathrm{w}}_1\right)].\frac{\mathrm{ch}\left({\mathrm{w}}_1\right)}{\mathrm{ch}\left(2^{\mathrm{n}-1}{\mathrm{w}}_1\right)}$$$$=(1+{\mathrm{p}}_1)$$$$\mathrm{ch}\left(\mathrm{x}\right)\mathrm{sh}\left(\mathrm{x}\right)=\mathrm{sh}\left(2\mathrm{x}\right)\Rightarrow$$$$2^{\mathrm{n}-1}.\mathrm{sh}\left({\mathrm{w}}_1\right)\underset{\mathrm{k}=2}{\prod ^{\mathrm{n}}}\mathrm{ch}\left(2^{\mathrm{k}-2}{\mathrm{w}}_1\right)=\mathrm{sh}\left(2^{\mathrm{n}-1}{\mathrm{w}}_1\right)$$$$\mathrm{\Pi }\left(\mathrm{n}\right)=\frac{(1+{\mathrm{p}}_1).\mathrm{sh}\left(2^{\mathrm{n}-1}{\mathrm{w}}_1\right)}{\mathrm{ch}\left(2^{\mathrm{n}-1}{\mathrm{w}}_1\right).\mathrm{sh}\left({\mathrm{w}}_1\right)}=\frac{1+{\mathrm{p}}_1}{\mathrm{sh}\left({\mathrm{w}}_1\right)}.\mathrm{th}\left(2^{\mathrm{n}-1}{\mathrm{w}}_1\right)$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{\Pi }\left(\mathrm{n}\right)=\frac{1+{\mathrm{p}}_1}{\mathrm{sh}\left({\mathrm{w}}_1\right)}$$$$\mathrm{sh}\left({\mathrm{w}}_1\right)={({\mathrm{ch}}^2\left({\mathrm{w}}_1\right)-1)}^{\frac{1}{2}}=\sqrt{{\mathrm{x}}^2-1}$$$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\mathrm{\Pi }\left(\mathrm{n}\right)=\frac{1+\mathrm{x}}{\sqrt{{\mathrm{x}}^2-1}}=\sqrt{\frac{1+\mathrm{x}}{\mathrm{x}-1}}$$$$$$$$$$$$$$

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