Question Number 62656 by mathmax by abdo last updated on 24/Jun/19
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\:\:\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} } \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
$${let}\:{A}_{{n}} =\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} }\:\Rightarrow\:{A}_{{n}} =\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{{n}} \:×\frac{\mathrm{1}}{{n}}\:=\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} .\frac{\mathrm{1}}{{n}} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \:={e}^{{nln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)} \:\:\:\:\rightarrow{e}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} ={lim}_{{n}\rightarrow+\infty} \:\frac{{e}}{{n}}\:=\mathrm{0} \\ $$
Answered by MJS last updated on 24/Jun/19
$$\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} }=\frac{{n}^{{n}} }{{n}^{{n}+\mathrm{1}} }+{c}_{\mathrm{1}} \frac{{n}^{{n}−\mathrm{1}} }{{n}^{{n}+\mathrm{1}} }+{c}_{\mathrm{2}} \frac{{n}^{{n}−\mathrm{2}} }{{n}^{{n}+\mathrm{1}} }+...+{c}_{{n}} \frac{\mathrm{1}}{{n}^{{n}+\mathrm{1}} } \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{{n}^{{n}−{k}} }{{n}^{{n}+\mathrm{1}} }=\mathrm{0}\:\mathrm{for}\:\mathrm{0}\leqslant{k}\leqslant{n}\:\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right)^{{n}} }{{n}^{{n}+\mathrm{1}} }=\mathrm{0} \\ $$