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Let-u-n-be-a-set-satisfying-u-1-1-amp-u-n-1-u-n-ln-n-u-n-n-1-1-Prove-that-u-2023-gt-2023-ln-2023-2-Find-lim-n-u-n-ln-n-n-

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

Question-209129

Question Number 209129 by Adeyemi889 last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $$\frac{{x}^{\mathrm{3}} +\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}={x}+\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{{x}+\mathrm{1}}\Rightarrow{x}+\mathrm{3}=\left({A}+{B}\right){x}+{A}−{B}…

Question-209023

Question Number 209023 by Spillover last updated on 30/Jun/24 Commented by Spillover last updated on 01/Jul/24 $${let}\:{u}=\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\:\:\:\:\:\:\frac{{du}}{{dx}}=\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)=\frac{\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{1}+\mathrm{10}{x}+\mathrm{25}{x}^{\mathrm{2}} } \\…

Question-209030

Question Number 209030 by Spillover last updated on 30/Jun/24 Answered by aleks041103 last updated on 30/Jun/24 $${moment}\:{generating}\:{function}\:{is} \\ $$$${m}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}_{{k}} }{{k}!}{t}^{{k}} ={m}_{\mathrm{0}} +{m}_{\mathrm{1}} {t}+……