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Question Number 109766 by Karani last updated on 25/Aug/20
lim_(n→∞) ((2^(n/2) +(n+1)!)/(n(3^n +n!)))=?
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{\frac{{n}}{\mathrm{2}}} +\left({n}+\mathrm{1}\right)!}{{n}\left(\mathrm{3}^{{n}} +{n}!\right)}=? \\ $$
Answered by mathmax by abdo last updated on 25/Aug/20
U_n =((((√2))^n +(n+1)!)/(n3^n +nn!)) ⇒ U_n =(((n+1)!{1+((((√2))^n )/((n+1)!))})/(nn!{ (3^n /(n!))+1})) =((n+1)/n)×((1+((((√2))^n )/((n+1)!)))/(1+(3^n /(n!)))) we have ln((3^n /(n!))) =nln(3)−ln(n!) ∼nln(3)−ln(n^n e^(−n) (√(2πn))) =nln(3)−nln(n)−n−(1/2)ln(2πn) =n{ln3−ln(n)−1−(1/2)((ln(2πn))/n)}→−∞ ⇒lim_(n→+∞) (3^n /(n!)) =0 (n! defeat 3^n ..!) also we have lim_(n→+∞) ((((√2))^n )/((n+1)!)) =0 ⇒ ★lim_(n→+∞) U_n =1★
$$\mathrm{U}_{\mathrm{n}} =\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} \:+\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n3}^{\mathrm{n}} \:+\mathrm{nn}!}\:\Rightarrow\:\mathrm{U}_{\mathrm{n}} =\frac{\left(\mathrm{n}+\mathrm{1}\right)!\left\{\mathrm{1}+\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)!}\right\}}{\mathrm{nn}!\left\{\:\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}+\mathrm{1}\right\}} \\ $$$$=\frac{\mathrm{n}+\mathrm{1}}{\mathrm{n}}×\frac{\mathrm{1}+\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)!}}{\mathrm{1}+\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{ln}\left(\frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\right)\:=\mathrm{nln}\left(\mathrm{3}\right)−\mathrm{ln}\left(\mathrm{n}!\right) \\ $$$$\sim\mathrm{nln}\left(\mathrm{3}\right)−\mathrm{ln}\left(\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\right)\:=\mathrm{nln}\left(\mathrm{3}\right)−\mathrm{nln}\left(\mathrm{n}\right)−\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right) \\ $$$$=\mathrm{n}\left\{\mathrm{ln3}−\mathrm{ln}\left(\mathrm{n}\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{ln}\left(\mathrm{2}\pi\mathrm{n}\right)}{\mathrm{n}}\right\}\rightarrow−\infty\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \frac{\mathrm{3}^{\mathrm{n}} }{\mathrm{n}!}\:=\mathrm{0} \\ $$$$\left(\mathrm{n}!\:\mathrm{defeat}\:\mathrm{3}^{\mathrm{n}} \:\:..!\right)\:\:\mathrm{also}\:\mathrm{we}\:\mathrm{have}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)!}\:=\mathrm{0}\:\Rightarrow \\ $$$$\bigstar\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\:\mathrm{U}_{\mathrm{n}} =\mathrm{1}\bigstar \\ $$

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