Question Number 170659 by ArielVyny last updated on 28/May/22
$${c}\geqslant\mathrm{0}\:\left({U}_{{n}} \right)_{{n}\geqslant\mathrm{1}} \:\:\:{U}_{\mathrm{1}} =\mathrm{1} \\ $$$${U}_{{n}+\mathrm{1}} =\sqrt{{U}_{{n}} +{cn}} \\ $$$${show}\:{that}\:{U}_{{n}} \leqslant\alpha\sqrt{{n}} \\ $$$${determine}\:\alpha \\ $$