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

Given   { ((u_0 =a)),((u_(n+1) =a+(1/2)(1−(1/b^n ))u_n )) :}  a>0  b>1  a\  Calculate  u_1 ,  u_2 ,  and  u_3 .  b\  Show  that  the  sequence  (u_n )_(n∈N)   is  increasing.  c\  Deduce  that  (u_n )_(n∈N)   converges  and  determine  its  limit.

$$\mathcal{G}\mathrm{iven}\:\:\begin{cases}{\mathrm{u}_{\mathrm{0}} =\mathrm{a}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{n}} }\right)\mathrm{u}_{\mathrm{n}} }\end{cases}\:\:\mathrm{a}>\mathrm{0}\:\:\mathrm{b}>\mathrm{1} \\ $$ $$\mathrm{a}\backslash\:\:\mathcal{C}\mathrm{alculate}\:\:\mathrm{u}_{\mathrm{1}} ,\:\:\mathrm{u}_{\mathrm{2}} ,\:\:\mathrm{and}\:\:\mathrm{u}_{\mathrm{3}} . \\ $$ $$\mathrm{b}\backslash\:\:\mathcal{S}\mathrm{how}\:\:\mathrm{that}\:\:\mathrm{the}\:\:\mathrm{sequence}\:\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}} \:\:\mathrm{is}\:\:\mathrm{increasing}. \\ $$ $$\mathrm{c}\backslash\:\:\mathcal{D}\mathrm{educe}\:\:\mathrm{that}\:\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}} \:\:\mathrm{converges}\:\:\mathrm{and}\:\:\mathrm{determine}\:\:\mathrm{its}\:\:\mathrm{limit}. \\ $$

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

(a)u_0  = a ⇒ u_1  = a + (1/2)(1−(1/b^0 ))u_0   ⇒ u_1  = a  u_2  = a + (1/2)(1 + (1/b))a = a[1 + (1/2)(1 − (1/b))]  u_3  =[ a + (1/2)(1−(1/b^2 ))]{a[1 + (1/2)(1−(1/b))]}  (b) n∈N and a > 0, b > 1  a [1 + (1/2)(1−(1/b))] > a for  a > 0 and b > 1  [ a + (1/2)(1−(1/b^2 ))]{a[1 + (1/2)(1−(1/b))]} > a[1 + (1/2)(1−(1/b))]   ⇒ the sequence is increasing for n ∈ N.  (c) (1/2)(1−(1/b^n ))u_n  < 1 ∀ n ∈ N ⇒ u_n  is convergent

$$\left(\mathrm{a}\right){u}_{\mathrm{0}} \:=\:{a}\:\Rightarrow\:{u}_{\mathrm{1}} \:=\:{a}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}^{\mathrm{0}} }\right){u}_{\mathrm{0}} \\ $$ $$\Rightarrow\:{u}_{\mathrm{1}} \:=\:{a} \\ $$ $${u}_{\mathrm{2}} \:=\:{a}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{b}}\right){a}\:=\:{a}\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{b}}\right)\right] \\ $$ $${u}_{\mathrm{3}} \:=\left[\:{a}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)\right]\left\{{a}\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}}\right)\right]\right\} \\ $$ $$\left(\mathrm{b}\right)\:{n}\in\mathbb{N}\:\mathrm{and}\:{a}\:>\:\mathrm{0},\:{b}\:>\:\mathrm{1} \\ $$ $${a}\:\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}}\right)\right]\:>\:{a}\:\mathrm{for}\:\:{a}\:>\:\mathrm{0}\:\mathrm{and}\:{b}\:>\:\mathrm{1} \\ $$ $$\left[\:{a}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)\right]\left\{{a}\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}}\right)\right]\right\}\:>\:{a}\left[\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}}\right)\right]\: \\ $$ $$\Rightarrow\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{for}\:{n}\:\in\:\mathbb{N}. \\ $$ $$\left(\mathrm{c}\right)\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{b}^{{n}} }\right){u}_{{n}} \:<\:\mathrm{1}\:\forall\:{n}\:\in\:\mathbb{N}\:\Rightarrow\:{u}_{{n}} \:\mathrm{is}\:\mathrm{convergent} \\ $$ $$ \\ $$

Commented byAr Brandon last updated on 01/Jun/20

Wow, so fast ! Thank you ��

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