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Question Number 144622 by phally last updated on 27/Jun/21

Answered by alcohol last updated on 27/Jun/21

$$e^{\frac{1}{2a}}$$

Answered by mathmax by abdo last updated on 27/Jun/21

$${\mathrm{U}}_{\mathrm{n}}=\prod _{\mathrm{k}=1}^{\mathrm{n}}(1+\frac{\mathrm{ka}}{{\mathrm{n}}^2})\Rightarrow {\mathrm{logU}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\log (1+\frac{\mathrm{ka}}{{\mathrm{n}}^2})\mathrm{we}\mathrm{have}$$$$\log (1+\frac{\mathrm{ka}}{{\mathrm{n}}^2})\sim \frac{\mathrm{ka}}{{\mathrm{n}}^2}(\mathrm{n}\to +\mathrm{\infty })\Rightarrow \mathrm{\Sigma }\log (...)\sim \sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{ka}}{{\mathrm{n}}^2}$$$$=\frac{\mathrm{a}}{{\mathrm{n}}^2}\frac{\mathrm{n}(\mathrm{n}+1)}{2}\Rightarrow {\mathrm{logU}}_{\mathrm{n}}\to \frac{\mathrm{a}}{2}\Rightarrow {\mathrm{U}}_{\mathrm{n}}\to {\mathrm{e}}^{\frac{\mathrm{a}}{2}}$$

Commented by phally last updated on 27/Jun/21

$$thankbrother$$

Commented by mathmax by abdo last updated on 27/Jun/21

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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