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Question Number 199309 by universe last updated on 01/Nov/23

a_(n+1 ) = a_n + (√(a_n ^2 + 1)) , a_0 = 0 find a_(n ) = ??

$$\:\:\:\:{a}_{{n}+\mathrm{1}\:} =\:{a}_{{n}} \:+\:\sqrt{{a}_{{n}} ^{\mathrm{2}} \:+\:\mathrm{1}}\:\:,\:{a}_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}\:} \:=\:\:?? \\ $$

Answered by TheHoneyCat last updated on 01/Nov/23

a_n ^2 >0 ⇒a_n ^2 +1>1 ⇒(√(a_n ^2 +1))>1 ⇒a_(n+1) >1 Thus (a_n ) is strictly increasinng it is therefore either convergent to l∈R and l=l+(√(l^2 +1)) or l=+∞ the eqution on l gives l^2 =−1 wich is not possible in R thus: a_n →_(n→∞) ∞ _■

$${a}_{{n}} ^{\mathrm{2}} >\mathrm{0} \\ $$$$\Rightarrow{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}>\mathrm{1} \\ $$$$\Rightarrow\sqrt{{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}}>\mathrm{1} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} >\mathrm{1} \\ $$$$\mathrm{Thus}\:\left({a}_{{n}} \right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{increasinng} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{therefore}\:\mathrm{either}\:\mathrm{convergent}\:\mathrm{to}\:{l}\in\mathbb{R} \\ $$$$\mathrm{and}\:{l}={l}+\sqrt{{l}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{or}\:{l}=+\infty \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{eqution}\:\mathrm{on}\:{l}\:\mathrm{gives}\:{l}^{\mathrm{2}} =−\mathrm{1}\:\mathrm{wich}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{possible}\:\mathrm{in}\:\mathbb{R} \\ $$$$ \\ $$$$\mathrm{thus}:\:{a}_{{n}} \:\:\underset{{n}\rightarrow\infty} {\rightarrow}\:\:\infty\:\:\:_{\blacksquare} \\ $$

Commented by TheHoneyCat last updated on 01/Nov/23

woops��. I misread the question. sorry...

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