Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 73251 by aliesam last updated on 09/Nov/19

Answered by mind is power last updated on 09/Nov/19

$$\mathrm{E}\left({\mathrm{u}}_{\mathrm{n}}\right)⩽{\mathrm{u}}_{\mathrm{n}}<\mathrm{E}\left({\mathrm{u}}_{\mathrm{n}}\right)+1={\mathrm{U}}_{\mathrm{n}+1}$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{creasing}\mathrm{squances}$$$${\mathrm{U}}_{\mathrm{n}+2}=\mathrm{E}(\mathrm{E}\left({\mathrm{u}}_{\mathrm{n}}\right)+1)+1=\mathrm{E}\left({\mathrm{u}}_{\mathrm{n}}\right)+2=1+{\mathrm{U}}_{\mathrm{n}+1}$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}+2}-{\mathrm{U}}_{\mathrm{n}+1}=1$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}+1}-{\mathrm{U}}_{\mathrm{n}}=1\mathrm{\forall }\mathrm{n}\in {\mathbb{N}}^{\ast }$$$$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}({\mathrm{U}}_{\mathrm{k}+1}-{\mathrm{U}}_{\mathrm{k}})=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}1$$$$\Rightarrow {\mathrm{U}}_{\mathrm{n}}-{\mathrm{U}}_1=\mathrm{n}-1\Rightarrow {\mathrm{U}}_{\mathrm{n}}=(\mathrm{n}-1)+{\mathrm{U}}_1$$$$\frac{{\mathrm{U}}_{\mathrm{n}}}{\mathrm{n}}=\frac{(\mathrm{n}-1)}{\mathrm{n}}+\frac{{\mathrm{U}}_1}{\mathrm{n}}\to 1$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com