Question Number 88761 by M±th+et£s last updated on 12/Apr/20
$${if}\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}} \\ $$$${and}\:{b}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!\sqrt{{n}}}{\mathrm{4}^{{n}} \:\left({n}!\right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\mathrm{1} \\ $$$${find}\:\underset{{n}\rightarrow\infty} {{lim}b}_{{n}} =? \\ $$
Commented by M±th+et£s last updated on 12/Apr/20
$${i}\:{need}\:{a}\:{help}\:{please}.. \\ $$
Commented by abdomathmax last updated on 12/Apr/20
$${strling}\:{formilae}\:\:\:{n}!\:\sim\:{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\left(\mathrm{2}{n}\right)!\:\sim\:\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}} \:{e}^{−\mathrm{2}{n}} \sqrt{\mathrm{4}\pi{n}}\Rightarrow \\ $$$${b}_{{n}} \sim\frac{\left(\mathrm{2}{n}\right)^{\mathrm{2}{n}} {e}^{−\mathrm{2}{n}} \sqrt{\mathrm{4}\pi{n}}×\sqrt{{n}}}{\mathrm{4}^{{n}} {n}^{\mathrm{2}{n}} \:{e}^{−\mathrm{2}{n}} \left(\mathrm{2}\pi{n}\right)} \\ $$$$=\frac{\mathrm{2}\sqrt{\pi}{n}}{\left(\mathrm{2}\pi{n}\right)}\:=\frac{\mathrm{1}}{\sqrt{\pi}}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:{b}_{{n}} =\frac{\mathrm{1}}{\sqrt{\pi}} \\ $$
Commented by TANMAY PANACEA. last updated on 12/Apr/20
Commented by M±th+et£s last updated on 12/Apr/20
$${thanx}\:{sir}\:{god}\:{bless}\:{you} \\ $$