Question Number 132367 by Raxreedoroid last updated on 13/Feb/21
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{{n}} }{\Gamma\left({n}+\mathrm{1}\right)} \\ $$
Answered by TheSupreme last updated on 13/Feb/21
![if n∈N → Γ(n+1)=n! lim (x^n /(n!))=0 ∀x∈R_0 ^+ sup((x^n /(n!)))=∞ with x∈R_0 ^+ set A=[0,a]∈R_0 ^+ sup((x^n /(n!)))=(a^n /(n!)) lim_n sup(f_n (x))=0 ∀x∈A, ∀a∈R^+ convergenza puntuale in R_0 ^+ convergenza assoluta ∀x∈A, ∀a∈R^+](https://www.tinkutara.com/question/Q132370.png)
$${if}\:{n}\in\mathbb{N}\:\rightarrow\:\Gamma\left({n}+\mathrm{1}\right)={n}! \\ $$$$\mathrm{lim}\:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\mathrm{0}\:\forall{x}\in\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$$\mathrm{sup}\left(\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\right)=\infty\:{with}\:{x}\in\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$${set}\:{A}=\left[\mathrm{0},{a}\right]\in\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$${sup}\left(\frac{{x}^{{n}} }{{n}!}\right)=\frac{{a}^{{n}} }{{n}!} \\ $$$${lim}_{{n}} {sup}\left({f}_{{n}} \left({x}\right)\right)=\mathrm{0}\:\forall{x}\in{A},\:\forall{a}\in\mathbb{R}^{+} \\ $$$${convergenza}\:{puntuale}\:{in}\:\mathbb{R}_{\mathrm{0}} ^{+} \\ $$$${convergenza}\:{assoluta}\:\forall{x}\in{A},\:\forall{a}\in\mathbb{R}^{+} \\ $$