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Question Number 124106 by mathocean1 last updated on 30/Nov/20
U_(n ) is a sequence of real numbers   defined by U_0 =0 and for n ∈ N,   U_(n+1) =(√(U_n +6))  1. show that 0≤U_n ≤3.  2. show that U_(n ) is non−decreasing.  3. show that 3−U_(n+1) ≤((3−U_n )/3)  4.  Deduct that 0≤3−U_(n+1) ≤((1/3))^n
$${U}_{{n}\:} {is}\:{a}\:{sequence}\:{of}\:{real}\:{numbers}\: \\ $$$${defined}\:{by}\:{U}_{\mathrm{0}} =\mathrm{0}\:{and}\:{for}\:{n}\:\in\:\mathbb{N},\: \\ $$$${U}_{{n}+\mathrm{1}} =\sqrt{{U}_{{n}} +\mathrm{6}} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{0}\leqslant{U}_{{n}} \leqslant\mathrm{3}. \\ $$$$\mathrm{2}.\:{show}\:{that}\:{U}_{{n}\:} {is}\:{non}−{decreasing}. \\ $$$$\mathrm{3}.\:{show}\:{that}\:\mathrm{3}−{U}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{3}−{U}_{{n}} }{\mathrm{3}} \\ $$$$\mathrm{4}.\:\:{Deduct}\:{that}\:\mathrm{0}\leqslant\mathrm{3}−{U}_{{n}+\mathrm{1}} \leqslant\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \\ $$

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