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Question Number 133928 by mnjuly1970 last updated on 25/Feb/21

Answered by mathmax by abdo last updated on 25/Feb/21

2) u_n =(C_(2n) ^n )^(1/n) we have C_(2n) ^n =(((2n)!)/((n!)^2 )) ⇒ u_n =e^((1/n)log((((2n)!)/((n!)^2 )))) we have n!∼n^n e^(−n) (√(2πn)) ⇒(n!)^2 ∼n^(2n) e^(−2n) (2πn) and (2n)! ∼(2n)^(2n) e^(−2n) (√(4πn)) =2^(2n) .n^(2n) e^(−2n) (√(4πn)) ⇒(((2n)!)/((n!)^2 ))∼ ((2^(2n) .n^(2n) .e^(−2n) )/(n^(2n) e^(−2n) (2πn)))(√(4πn)) =2^(2n) ×(1/( (√(πn)))) ⇒log((((2n)!)/((n!)^2 )))∼2nlog(2)−(1/2)log(nπ) ⇒ (1/n)log(...) ∼2log(2)−((log(nπ))/(2n)) →log(4) ⇒lim_(n→+∞) u_n =e^(log(4)) =4

$$\left.\mathrm{2}\right)\:\mathrm{u}_{\mathrm{n}} =\left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:=\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{u}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}} }\right)} \:\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}} \\ $$$$\Rightarrow\left(\mathrm{n}!\right)^{\mathrm{2}} \:\sim\mathrm{n}^{\mathrm{2n}} \mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{n}\right)\:\:\mathrm{and}\:\left(\mathrm{2n}\right)!\:\sim\left(\mathrm{2n}\right)^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \sqrt{\mathrm{4}\pi\mathrm{n}} \\ $$$$=\mathrm{2}^{\mathrm{2n}} \:.\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \sqrt{\mathrm{4}\pi\mathrm{n}}\:\Rightarrow\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}} }\sim\:\frac{\mathrm{2}^{\mathrm{2n}} .\mathrm{n}^{\mathrm{2n}} .\mathrm{e}^{−\mathrm{2n}} }{\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{n}\right)}\sqrt{\mathrm{4}\pi\mathrm{n}} \\ $$$$=\mathrm{2}^{\mathrm{2n}} ×\frac{\mathrm{1}}{\:\sqrt{\pi\mathrm{n}}}\:\Rightarrow\mathrm{log}\left(\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}} }\right)\sim\mathrm{2nlog}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{n}\pi\right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{n}}\mathrm{log}\left(...\right)\:\sim\mathrm{2log}\left(\mathrm{2}\right)−\frac{\mathrm{log}\left(\mathrm{n}\pi\right)}{\mathrm{2n}}\:\rightarrow\mathrm{log}\left(\mathrm{4}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{e}^{\mathrm{log}\left(\mathrm{4}\right)} \:=\mathrm{4} \\ $$

Commented by mnjuly1970 last updated on 26/Feb/21

thanks alot sir max

$$\:\:\:{thanks}\:{alot}\:{sir}\:{max} \\ $$

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