Question Number 201495 by Rodier97 last updated on 07/Dec/23
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\left({Un}\right)_{{n}\geqslant\mathrm{1}\:;} \:\:\:\:\frac{\mathrm{1}}{{nC}_{\mathrm{2}{n}} ^{{n}} \:} \\ $$$$ \\ $$$$\:\:\:\:{study}\:\:{convergence} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Faetmaaa last updated on 07/Dec/23
$$\forall{n}\geqslant\mathrm{1}\:\:\:\:\mathrm{0}\:\leqslant\:{Un}\:\leqslant\:\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{So}\:\:\mathrm{lim}\:\left({Un}\right)_{{n}\geqslant\mathrm{1}} \:=\:\mathrm{0} \\ $$
Answered by MM42 last updated on 07/Dec/23
$${u}_{{n}} =\frac{\left({n}!\right)^{\mathrm{2}} }{{n}×\left(\mathrm{2}{n}\right)!}\:\:\:\:;\:\:\:{u}_{{n}} >\mathrm{0}\:\:\:\left({i}\right) \\ $$$${u}_{{n}+\mathrm{1}} =\frac{\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)×\left(\mathrm{2}{n}+\mathrm{2}\right)!}\:\:\: \\ $$$$ \\ $$$$\frac{{u}_{{n}+\mathrm{1}} }{{u}_{{n}} }=\frac{\left(\left({n}+\mathrm{1}\right)!\right)^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)×\left(\mathrm{2}{n}+\mathrm{2}\right)!}×\frac{{n}×\left(\mathrm{2}{n}\right)!}{\left(\left({n}!\right)\right)^{\mathrm{2}} }\:\:\: \\ $$$$=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} ×\left(\left({n}!\right)\right)^{\mathrm{2}} }{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} ×\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}×\frac{{n}×\left(\mathrm{2}{n}\right)!}{\left(\left({n}!\right)\right)^{\mathrm{2}} }\: \\ $$$$=\frac{{n}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:<\mathrm{1} \\ $$$$\Rightarrow{u}_{{n}} \:\:{is}\:{decrasing}\:\:\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\rightarrow{u}_{{n}} \:\:{is}\:\:{vonvergence} \\ $$$$ \\ $$