Question Number 209631 by York12 last updated on 17/Jul/24
$$\mathrm{Let}\:{u}_{{n}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{set}\:\mathrm{satisfying}\:{u}_{\mathrm{1}} =\mathrm{1}\:\&\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} +\frac{\mathrm{ln}\:{n}}{{u}_{{n}} }\:\:,\:\forall\:{n}\:\geqslant\mathrm{1} \\ $$$$\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:{u}_{\mathrm{2023}} >\sqrt{\mathrm{2023}.\mathrm{ln}\:\mathrm{2023}}. \\ $$$$\mathrm{2}.\:\mathrm{Find}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{u}_{{n}} .\mathrm{ln}\:{n}}{{n}}. \\ $$