Question Number 196164 by Rodier97 last updated on 19/Aug/23
$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{calcul}: \\ $$$$ \\ $$$$\:\:\:{lim}_{{n}\rightarrow+\infty\:\:\:\:} {Un}=\sqrt[{{n}}]{\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}^{{n}} }} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by universe last updated on 19/Aug/23
$$\mathrm{4}/{e}\:\:\:{use}\:{stirling}\:{approximation} \\ $$
Answered by MM42 last updated on 20/Aug/23
![lnu_n =(1/n)((((2n)!)/(n!n^n )))=(1/n)ln(((2n(2n−1)...(n+1))/n^n )) =(1/n)Σ_(i=1) ^n ln(2−((i−1)/n)) ⇒lim_(n→∞) u_n =∫_0 ^1 ln(2−x)dx=(x−2)ln(2−x)−x]_0 ^1 =−1+2ln2 ⇒u_n =e^(−1+2ln2) =(4/e) ✓](https://www.tinkutara.com/question/Q196240.png)
$${lnu}_{{n}} =\frac{\mathrm{1}}{{n}}\left(\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}^{{n}} }\right)=\frac{\mathrm{1}}{{n}}{ln}\left(\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)…\left({n}+\mathrm{1}\right)}{{n}^{{n}} }\right) \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{2}−\frac{{i}−\mathrm{1}}{{n}}\right) \\ $$$$\left.\Rightarrow{lim}_{{n}\rightarrow\infty} \:{u}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{2}−{x}\right){dx}=\left({x}−\mathrm{2}\right){ln}\left(\mathrm{2}−{x}\right)−{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =−\mathrm{1}+\mathrm{2}{ln}\mathrm{2} \\ $$$$\Rightarrow{u}_{{n}} ={e}^{−\mathrm{1}+\mathrm{2}{ln}\mathrm{2}} =\frac{\mathrm{4}}{{e}}\:\:\checkmark \\ $$$$ \\ $$$$ \\ $$$$ \\ $$