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Question Number 149795 by mathdanisur last updated on 07/Aug/21
lim_(n→∞) ((1∙3∙5∙7∙ ... ∙(2n-1))/(2∙4∙6∙ ... ∙2n)) = ?
$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}\centerdot\mathrm{3}\centerdot\mathrm{5}\centerdot\mathrm{7}\centerdot\:…\:\centerdot\left(\mathrm{2n}-\mathrm{1}\right)}{\mathrm{2}\centerdot\mathrm{4}\centerdot\mathrm{6}\centerdot\:…\:\centerdot\mathrm{2n}}\:=\:? \\ $$
Answered by mathmax by abdo last updated on 07/Aug/21
u_n =((1.3.5.7.....(2n−1))/(2.4.6....(2n))) =((1.3.5....(2n−1))/(2^n n!)) =((1.2.3.4.5.....(2n−1).2n)/((2^n )^2 (n!)^2 ))=(((2n)!)/(2^(2n) (n!)^2 )) ⇒ we have n!∼n^n e^(−n) (√(2π))n and (2n)! ∼(2n)^(2n) e^(−2n) (√(2π(2n)))=2^(2n) n^(2n) e^(−2n) (2(√(πn))) ⇒ u_n ∼((2^(2n) n^(2n) e^(−2n) (2(√(πn))))/(2^(2n) n^(2n) e^(−2n) (2πn))) ⇒u_n ∼(1/( (√(nπ)))) ⇒lim_(n→+∞) u_n =0
$$\mathrm{u}_{\mathrm{n}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}…..\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}….\left(\mathrm{2n}\right)}\:=\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2n}−\mathrm{1}\right)}{\mathrm{2}^{\mathrm{n}} \:\mathrm{n}!} \\ $$$$=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…..\left(\mathrm{2n}−\mathrm{1}\right).\mathrm{2n}}{\left(\mathrm{2}^{\mathrm{n}} \right)^{\mathrm{2}} \left(\mathrm{n}!\right)^{\mathrm{2}} }=\frac{\left(\mathrm{2n}\right)!}{\mathrm{2}^{\mathrm{2n}} \left(\mathrm{n}!\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{n}!\sim\mathrm{n}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi}\mathrm{n}\:\mathrm{and} \\ $$$$\left(\mathrm{2n}\right)!\:\sim\left(\mathrm{2n}\right)^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \sqrt{\mathrm{2}\pi\left(\mathrm{2n}\right)}=\mathrm{2}^{\mathrm{2n}} \:\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\sqrt{\pi\mathrm{n}}\right)\:\Rightarrow \\ $$$$\mathrm{u}_{\mathrm{n}} \sim\frac{\mathrm{2}^{\mathrm{2n}} \:\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\sqrt{\pi\mathrm{n}}\right)}{\mathrm{2}^{\mathrm{2n}} \:\mathrm{n}^{\mathrm{2n}} \:\mathrm{e}^{−\mathrm{2n}} \left(\mathrm{2}\pi\mathrm{n}\right)}\:\Rightarrow\mathrm{u}_{\mathrm{n}} \sim\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}\pi}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} =\mathrm{0} \\ $$
Commented by mathdanisur last updated on 07/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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