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Un-n-1-1-nC-2n-n-study-convergence-




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

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