Question Number 29452 by prof Abdo imad last updated on 08/Feb/18
$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:\prod_{{k}=\mathrm{1}} ^{{n}} \:\:\left(\mathrm{1}\:+\frac{{k}}{{n}}\right)^{\frac{\mathrm{1}}{{n}}} \:\:. \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
![let put A_n =Π_(k=1) ^n (1+(k/n))^(1/n) ⇒ ln(A_n )=(1/n) Σ_(k=1) ^n ln(1 + (k/n)) ⇒ lim_(n→∞) ln(A_n )=lim_(n→∞) ((1−0)/n) Σ_(k=1) ^n ln(1 +((k(1−0))/n)) =∫_0 ^1 ln(1+x)dx=_(1+x=t) ∫_1 ^2 ln(t)dt =[tlnt −t]_1 ^2 = (2ln2 −2) −(−1)= 2ln2−1. ⇒lim_(n→∞) A_n = e^(ln(4)−1) = (4/e) .](https://www.tinkutara.com/question/Q30398.png)
$${let}\:{put}\:\:{A}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}}\right)^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow \\ $$$${ln}\left({A}_{{n}} \right)=\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}\:+\:\frac{{k}}{{n}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:{ln}\left({A}_{{n}} \right)={lim}_{{n}\rightarrow\infty} \:\:\frac{\mathrm{1}−\mathrm{0}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{ln}\left(\mathrm{1}\:+\frac{{k}\left(\mathrm{1}−\mathrm{0}\right)}{{n}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{ln}\left(\mathrm{1}+{x}\right){dx}=_{\mathrm{1}+{x}={t}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:{ln}\left({t}\right){dt} \\ $$$$=\left[{tlnt}\:−{t}\right]_{\mathrm{1}} ^{\mathrm{2}} =\:\left(\mathrm{2}{ln}\mathrm{2}\:−\mathrm{2}\right)\:−\left(−\mathrm{1}\right)=\:\mathrm{2}{ln}\mathrm{2}−\mathrm{1}. \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \:{A}_{{n}} =\:{e}^{{ln}\left(\mathrm{4}\right)−\mathrm{1}} =\:\frac{\mathrm{4}}{{e}}\:\:. \\ $$