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Question-221260

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Question Number 221260 by mnjuly1970 last updated on 28/May/25
Answered by maths2 last updated on 30/May/25
=lim_(n→∞) (√n).Π_(k=1) ^n ((2^(2n) (n!)^2 )/((2n+1)!)) n!∼(√(2πn)).((n/e))^n .. lim_(n→∞) (√n).(((2πn)((n/e))^(2n) )/((2π(2n+1))(((2n+1)/e))^(2n+1) )) =lim_(n→∞) (√(n.))(1/(2n+1)).e.(n^(2n+1) /((2n+1))) ((n/(2n+1)))^(2n+1) =e^((2n+1)ln((n/(2n+1)))) =e^((2n+1)ln((1/2)+−(1/(2(2n+1))))) =e^((2n+1){−ln(2)+ln(1−(1/(2n+1))))) =e^((2n+1){−ln(2)−(1/((2n+1)))−(1/(2(2n+1)^2 ))+o((1/((2n+1)^2 )))}) =(√(2n+1)){e^(−1−(1/(2(2n+1)))+o((1/(2n+1)))) } =(√(2n+1)).(1/e){1−(1/(2(2n+1)))+o((1/(2n+1)))}∼((√(2n+1))/e) =(√n).(e/((2n+1)))..((√(2n+1))/e)=(√2)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}.\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${n}!\sim\sqrt{\mathrm{2}\pi{n}}.\left(\frac{{n}}{{e}}\right)^{{n}} .. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}.\frac{\left(\mathrm{2}\pi{n}\right)\left(\frac{{n}}{{e}}\right)^{\mathrm{2}{n}} }{\left(\mathrm{2}\pi\left(\mathrm{2}{n}+\mathrm{1}\right)\right)\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{e}}\right)^{\mathrm{2}{n}+\mathrm{1}} } \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}.}\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}.{e}.\frac{{n}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}{n}+\mathrm{1}} ={e}^{\left(\mathrm{2}{n}+\mathrm{1}\right){ln}\left(\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right)} ={e}^{\left(\mathrm{2}{n}+\mathrm{1}\right){ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}\right)} \\ $$$$={e}^{\left(\mathrm{2}{n}+\mathrm{1}\right)\left\{−{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\right)} \\ $$$$={e}^{\left(\mathrm{2}{n}+\mathrm{1}\right)\left\{−{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+{o}\left(\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\right\}} \\ $$$$=\sqrt{\mathrm{2}{n}+\mathrm{1}}\left\{{e}^{−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}+{o}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)} \right\} \\ $$$$=\sqrt{\mathrm{2}{n}+\mathrm{1}}.\frac{\mathrm{1}}{{e}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)}+{o}\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)\right\}\sim\frac{\sqrt{\mathrm{2}{n}+\mathrm{1}}}{{e}} \\ $$$$=\sqrt{{n}}.\frac{{e}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}..\frac{\sqrt{\mathrm{2}{n}+\mathrm{1}}}{{e}}=\sqrt{\mathrm{2}} \\ $$

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