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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^(1/(2a))

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

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

U_n =Π_(k=1) ^n (1+((ka)/n^2 )) ⇒logU_n =Σ_(k=1) ^n log(1+((ka)/n^2 )) we have log(1+((ka)/n^2 ))∼((ka)/n^2 )(n→+∞) ⇒Σlog(...)∼Σ_(k=1) ^n ((ka)/n^2 ) =(a/n^2 )((n(n+1))/2) ⇒logU_n →(a/2) ⇒U_n →e^(a/2)

$$\mathrm{U}_{\mathrm{n}} =\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \left(\mathrm{1}+\frac{\mathrm{ka}}{\mathrm{n}^{\mathrm{2}} }\right)\:\Rightarrow\mathrm{logU}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{ka}}{\mathrm{n}^{\mathrm{2}} }\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{ka}}{\mathrm{n}^{\mathrm{2}} }\right)\sim\frac{\mathrm{ka}}{\mathrm{n}^{\mathrm{2}} }\left(\mathrm{n}\rightarrow+\infty\right)\:\Rightarrow\Sigma\mathrm{log}\left(...\right)\sim\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{ka}}{\mathrm{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{a}}{\mathrm{n}^{\mathrm{2}} }\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\:\Rightarrow\mathrm{logU}_{\mathrm{n}} \rightarrow\frac{\mathrm{a}}{\mathrm{2}}\:\Rightarrow\mathrm{U}_{\mathrm{n}} \rightarrow\mathrm{e}^{\frac{\mathrm{a}}{\mathrm{2}}} \\ $$

Commented by phally last updated on 27/Jun/21

thank brother

$${thank}\:{brother} \\ $$

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

you are welcome

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

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