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Question Number 65778 by mathmax by abdo last updated on 03/Aug/19
$${find}\:{lim}_{{n}\rightarrow+\infty} \:{e}^{−{n}^{\mathrm{2}} } \left({n}+\mathrm{1}\right)^{{n}!} \\ $$
Commented by mathmax by abdo last updated on 08/Aug/19
$${let}\:{A}_{{n}} ={e}^{−{n}^{\mathrm{2}} } \left({n}+\mathrm{1}\right)^{{n}!} \:\Rightarrow{A}_{{n}} ={e}^{−{n}^{\mathrm{2}} } \:{e}^{{n}!{ln}\left({n}+\mathrm{1}\right)} \\ $$$$=\:{e}^{−{n}^{\mathrm{2}} \:+{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)!{ln}\left({n}+\mathrm{1}\right)} \:={e}^{{n}^{\mathrm{2}} \left(−\mathrm{1}\:+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left({n}−\mathrm{2}\right)!{ln}\left({n}+\mathrm{1}\right)\right)} \\ $$$${its}\:{clear}\:{that}\:{lim}_{{n}\rightarrow+\infty} \:−\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\left({n}−\mathrm{2}\right)!{ln}\left({n}+\mathrm{1}\right)\:=+\infty\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty \\ $$