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Question Number 204929 by universe last updated on 02/Mar/24

$$\underset{n\to \mathrm{\infty }}{\lim }\underset{r=1}{\prod ^n}\frac{n^2-r}{n^2+r}=?$$

Answered by witcher3 last updated on 02/Mar/24

$$\underset{\mathrm{r}=1}{\prod ^{\mathrm{n}}}\frac{({\mathrm{n}}^2-\mathrm{r})}{({\mathrm{n}}^2+\mathrm{r})}=\frac{{\mathrm{n}}^2({\mathrm{n}}^2-1)....({\mathrm{n}}^2-\mathrm{n})}{({\mathrm{n}}^2+\mathrm{n})....\left({\mathrm{n}}^2\right)}=\frac{{\mathrm{n}}^2!({\mathrm{n}}^2-1)!}{({\mathrm{n}}^2-\mathrm{n}-1)!.({\mathrm{n}}^2+\mathrm{n})!}={\mathrm{U}}_{\mathrm{n}}$$$$\mathrm{n}!\sim \sqrt{2\pi \mathrm{n}}.{\left(\frac{\mathrm{n}}{\mathrm{e}}\right)}^{\mathrm{n}}$$$${\mathrm{U}}_{\mathrm{n}}\sim \frac{\sqrt{2\pi {\mathrm{n}}^2}.{\left(\frac{{\mathrm{n}}^2}{\mathrm{e}}\right)}^{{\mathrm{n}}^2}.\sqrt{2\pi ({\mathrm{n}}^2-1)}.{\left(\frac{{\mathrm{n}}^2-1}{\mathrm{e}}\right)}^{{\mathrm{n}}^2-1}}{\sqrt{2\pi ({\mathrm{n}}^2-\mathrm{n}-1)}.{\left(\frac{{\mathrm{n}}^2-\mathrm{n}-1}{\mathrm{e}}\right)}^{{\mathrm{n}}^2-\mathrm{n}-1}.\sqrt{2\pi ({\mathrm{n}}^2+\mathrm{n})}.{\left(\frac{{\mathrm{n}}^2+\mathrm{n}}{\mathrm{e}}\right)}^{{\mathrm{n}}^2+\mathrm{n}}}={\mathrm{W}}_{\mathrm{n}}$$$$\text{Double subscripts: use braces to clarify}$$$${\mathrm{W}}_{\mathrm{n}}=\sqrt{\frac{4{\pi }^2{\mathrm{n}}^2.({\mathrm{n}}^2-1)}{4{\pi }^2({\mathrm{n}}^2-\mathrm{n}-1)({\mathrm{n}}^2+\mathrm{n})}}.\frac{{({\mathrm{n}}^2-1)}^{{\mathrm{n}}^2-1}.{\mathrm{n}}^{{2\mathrm{n}}^2}}{{({\mathrm{n}}^2+\mathrm{n})}^{{\mathrm{n}}^2+\mathrm{n}}{({\mathrm{n}}^2-\mathrm{n}-1)}^{{\mathrm{n}}^2-\mathrm{n}-1}}.\frac{{\mathrm{e}}^{-{\mathrm{n}}^2-{\mathrm{n}}^2+1}}{{\mathrm{e}}^{-{\mathrm{n}}^2+\mathrm{n}+1+{\mathrm{n}}^2-\mathrm{n}}}$$$${\mathrm{W}}_{\mathrm{n}}\sim \frac{{({\mathrm{n}}^2-1)}^{{\mathrm{n}}^2-1}{\mathrm{n}}^{{2\mathrm{n}}^2}}{{({\mathrm{n}}^2+\mathrm{n})}^{{\mathrm{n}}^2+\mathrm{n}}{({\mathrm{n}}^2-\mathrm{n}-1)}^{{\mathrm{n}}^2-\mathrm{n}-1}}$$$${\mathrm{e}}^{({\mathrm{n}}^2-1)\ln ({\mathrm{n}}^2-1)}=\frac{{\mathrm{e}}^{{\mathrm{n}}^2\ln \left({\mathrm{n}}^2\right)+{\mathrm{n}}^2\left(\ln (1-\frac{1}{{\mathrm{n}}^2})\right)}.{\mathrm{n}}^{2\mathrm{n}}}{{\mathrm{e}}^{({\mathrm{n}}^2+\mathrm{n})\ln ({\mathrm{n}}^2+\mathrm{n})}.{\mathrm{e}}^{({\mathrm{n}}^2-\mathrm{n})\ln ({\mathrm{n}}^2-\mathrm{n}-1)}}$$$${\mathrm{W}}_{\mathrm{n}}\sim \frac{{\mathrm{n}}^{4\mathrm{n}}.{\mathrm{e}}^{{\mathrm{n}}^2\ln (1-\frac{1}{{\mathrm{n}}^2})}}{{\mathrm{e}}^{+\mathrm{nln}(1+\frac{1}{\mathrm{n}})+{\mathrm{n}}^2\ln (1+\frac{1}{\mathrm{n}})}.{\mathrm{e}}^{{\mathrm{n}}^2\ln (1-\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^2})-\mathrm{nln}(1-\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^2})}.{\mathrm{n}}^{4\mathrm{n}}}$$$$\ln (1+\mathrm{x})=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\mathrm{o}\left({\mathrm{x}}^2\right)$$$${\mathrm{Y}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{nln}(1+\frac{1}{\mathrm{n}})-\mathrm{nln}(1-\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^2})}={\mathrm{e}}^{\mathrm{n}(\frac{1}{\mathrm{n}}-\frac{1}{{2\mathrm{n}}^2})-\mathrm{n}(-(\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})-\frac{1}{2}{(\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})}^2+\mathrm{o}\left(\frac{1}{{\mathrm{n}}^2}\right)}$$$$\text{Double subscripts: use braces to clarify}$$$${\mathrm{Z}}_{\mathrm{n}}={\mathrm{e}}^{{\mathrm{n}}^2\ln (1+\frac{1}{\mathrm{n}})+{\mathrm{n}}^2\ln (1-\frac{1}{\mathrm{n}}-\frac{1}{{\mathrm{n}}^2})}={\mathrm{e}}^{\mathrm{n}(1-\frac{1}{2\mathrm{n}})+{\mathrm{n}}^2(-(\frac{1}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})-\frac{1}{{2\mathrm{n}}^2})+\mathrm{o}\left({\mathrm{n}}^2\right)}$$$${\mathrm{Z}}_{\mathrm{n}}\to {\mathrm{e}}^{-2}$$$$\underset{x\to \mathrm{\infty }}{\lim }{\mathrm{W}}_{\mathrm{x}}=\frac{{\mathrm{e}}^{-1}}{{\mathrm{e}}^2.{\mathrm{e}}^{-2}}=\frac{1}{\mathrm{e}}$$$$$$

Commented by universe last updated on 03/Mar/24

$$thankusir$$

Commented by witcher3 last updated on 04/Mar/24

$$\mathrm{Withe}\mathrm{Pleasur}$$

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