Question Number 151889 by mathdanisur last updated on 23/Aug/21
$$\mathrm{Compare}: \\ $$$$\sqrt[{\mathrm{2020}}]{\left(\mathrm{2020}!\right)^{\mathrm{3}} }\:\:\:\mathrm{and}\:\:\:\mathrm{505}\centerdot\mathrm{2021}^{\mathrm{2}} \\ $$
Answered by MJS_new last updated on 24/Aug/21
$$\forall{n}\in\mathbb{N}\mid{n}>\mathrm{1}:\left({n}!\right)^{\frac{\mathrm{3}}{{n}}} <\frac{{n}}{\mathrm{4}}×\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$