Question Number 143491 by SOMEDAVONG last updated on 15/Jun/21
$$\mathrm{Find}\:\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\mathrm{u}_{\mathrm{n}} \right),\mathrm{If}\:\begin{cases}{\mathrm{u}_{\mathrm{0}} =\mathrm{1},\mathrm{n}=\mathrm{1},\mathrm{2},\mathrm{3},…..}\\{\mathrm{u}_{\mathrm{n}} =\:\frac{\mathrm{2018}}{\mathrm{2019}}\mathrm{u}_{\mathrm{n}−\mathrm{1}} +\:\frac{\mathrm{1}}{\left(\mathrm{u}_{\mathrm{n}−\mathrm{1}} \right)^{\mathrm{2018}} }}\end{cases} \\ $$
Commented by mr W last updated on 15/Jun/21
$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{u}_{{n}} \\ $$$${L}=\frac{\mathrm{2018}}{\mathrm{2019}}{L}+\frac{\mathrm{1}}{{L}^{\mathrm{2018}} } \\ $$$${L}^{\mathrm{2019}} =\mathrm{2019} \\ $$$${L}=\sqrt[{\mathrm{2019}}]{\mathrm{2019}} \\ $$