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Question Number 91631 by mathmax by abdo last updated on 02/May/20
$${find}\:{a}\:{equivalent}\:{of}\:\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left({k}\right) \\ $$
Commented by mathmax by abdo last updated on 02/May/20
$${U}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:{ln}\left({k}\right)\:\Rightarrow{U}_{{n}} ={ln}\left(\prod_{{k}=\mathrm{2}} ^{{n}} \:{k}\right)\:={ln}\left({n}!\right)\:{we}\:{have} \\ $$$${n}!\:\sim{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow{ln}\left({n}!\right)\sim{nln}\left({n}\right)−{n}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\pi\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)\:\Rightarrow \\ $$$${U}_{{n}} \sim{nln}\left({n}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{ln}\left({n}\right)}+\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\:\Rightarrow{U}_{{n}} \sim\:{nln}\left({n}\right)\:\left({n}\rightarrow\infty\right) \\ $$