Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 209631 by York12 last updated on 17/Jul/24

$$\mathrm{Let}{u}_n\mathrm{be}\mathrm{a}\mathrm{set}\mathrm{satisfying}{u}_1=1\mathrm{\&}{u}_{n+1}=u_n+\frac{\ln n}{u_n},\mathrm{\forall }n⩾1$$$$1.\mathrm{Prove}\mathrm{that}{u}_{2023}>\sqrt{2023.\ln 2023}.$$$$2.\mathrm{Find}:\underset{n\to \mathrm{\infty }}{\lim }\frac{u_n.\ln n}{n}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com