Question Number 199309 by universe last updated on 01/Nov/23
$$\:\:\:\:{a}_{{n}+\mathrm{1}\:} =\:{a}_{{n}} \:+\:\sqrt{{a}_{{n}} ^{\mathrm{2}} \:+\:\mathrm{1}}\:\:,\:{a}_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:\mathrm{find}\:\mathrm{a}_{\mathrm{n}\:} \:=\:\:?? \\ $$
Answered by TheHoneyCat last updated on 01/Nov/23
$${a}_{{n}} ^{\mathrm{2}} >\mathrm{0} \\ $$$$\Rightarrow{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}>\mathrm{1} \\ $$$$\Rightarrow\sqrt{{a}_{{n}} ^{\mathrm{2}} +\mathrm{1}}>\mathrm{1} \\ $$$$\Rightarrow{a}_{{n}+\mathrm{1}} >\mathrm{1} \\ $$$$\mathrm{Thus}\:\left({a}_{{n}} \right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{increasinng} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{therefore}\:\mathrm{either}\:\mathrm{convergent}\:\mathrm{to}\:{l}\in\mathbb{R} \\ $$$$\mathrm{and}\:{l}={l}+\sqrt{{l}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{or}\:{l}=+\infty \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{eqution}\:\mathrm{on}\:{l}\:\mathrm{gives}\:{l}^{\mathrm{2}} =−\mathrm{1}\:\mathrm{wich}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{possible}\:\mathrm{in}\:\mathbb{R} \\ $$$$ \\ $$$$\mathrm{thus}:\:{a}_{{n}} \:\:\underset{{n}\rightarrow\infty} {\rightarrow}\:\:\infty\:\:\:_{\blacksquare} \\ $$
Commented by TheHoneyCat last updated on 01/Nov/23
woops��. I misread the question. sorry...