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 96489 by Ar Brandon last updated on 01/Jun/20

$$\mathcal{G}\mathrm{iven}\{\begin{array}{l}{\mathrm{u}}_0=\mathrm{a}\\ {\mathrm{u}}_{\mathrm{n}+1}=\mathrm{a}+\frac{1}{2}(1-\frac{1}{{\mathrm{b}}^{\mathrm{n}}}){\mathrm{u}}_{\mathrm{n}}\end{array}\mathrm{a}>0\mathrm{b}>1$$ $$\mathrm{a}\mathrm{\setminus }\mathcal{C}\mathrm{alculate}{\mathrm{u}}_1,{\mathrm{u}}_2,\mathrm{and}{\mathrm{u}}_3.$$ $$\mathrm{b}\mathrm{\setminus }\mathcal{S}\mathrm{how}\mathrm{that}\mathrm{the}\mathrm{sequence}{\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}\mathrm{is}\mathrm{increasing}.$$ $$\mathrm{c}\mathrm{\setminus }\mathcal{D}\mathrm{educe}\mathrm{that}{\left({\mathrm{u}}_{\mathrm{n}}\right)}_{\mathrm{n}\in \mathbb{N}}\mathrm{converges}\mathrm{and}\mathrm{determine}\mathrm{its}\mathrm{limit}.$$

Answered by Rio Michael last updated on 01/Jun/20

$$\left(\mathrm{a}\right)u_0=a\Rightarrow u_1=a+\frac{1}{2}(1-\frac{1}{b^0})u_0$$ $$\Rightarrow u_1=a$$ $$u_2=a+\frac{1}{2}(1+\frac{1}{b})a=a[1+\frac{1}{2}(1-\frac{1}{b})]$$ $$u_3=[a+\frac{1}{2}(1-\frac{1}{b^2})]\left\{a[1+\frac{1}{2}(1-\frac{1}{b})]\right\}$$ $$\left(\mathrm{b}\right)n\in \mathbb{N}\mathrm{and}a>0,b>1$$ $$a[1+\frac{1}{2}(1-\frac{1}{b})]>a\mathrm{for}a>0\mathrm{and}b>1$$ $$[a+\frac{1}{2}(1-\frac{1}{b^2})]\left\{a[1+\frac{1}{2}(1-\frac{1}{b})]\right\}>a[1+\frac{1}{2}(1-\frac{1}{b})]$$ $$\Rightarrow \mathrm{the}\mathrm{sequence}\mathrm{is}\mathrm{increasing}\mathrm{for}n\in \mathbb{N}.$$ $$\left(\mathrm{c}\right)\frac{1}{2}(1-\frac{1}{b^n})u_n<1\mathrm{\forall }n\in \mathbb{N}\Rightarrow u_n\mathrm{is}\mathrm{convergent}$$ $$$$

Commented byAr Brandon last updated on 01/Jun/20

Wow, so fast ! Thank you ��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com