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Question-151673




Question Number 151673 by mathdanisur last updated on 22/Aug/21
Answered by Kamel last updated on 22/Aug/21
L=lim_(n→+∞) Π_(k=n) ^(2n) (π/(π−Arctan((1/k))))=lim_(n→+∞) Π_(k=n) ^(2n) ((πk)/(πk−1))     =lim_(n→+∞) ((n.(n+1)(n+2)...(2n))/((n−(1/π))(n+1−(1/π))...(2n−(1/π))))     =lim_(n→+∞) ((Γ(2n+1)Γ(n−(1/π)))/(Γ(2n−(1/π)+1)Γ(n)))=lim_(n→+∞) (((n)^(2n) (n−(1/π)−1)^n 4^(1/π) )/((n−1)^n (n−(1/(2π)))^(2n) ))    =lim_(n→+∞) (((1−(1/(π(n−1))))^(n−1) 4^(1/π) )/((1−(1/(2πn)))^(2n) ))=(4)^(1/π)
$${L}=\underset{{n}\rightarrow+\infty} {{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\prod}}\frac{\pi}{\pi−{Arctan}\left(\frac{\mathrm{1}}{{k}}\right)}=\underset{{n}\rightarrow+\infty} {{lim}}\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\prod}}\frac{\pi{k}}{\pi{k}−\mathrm{1}} \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty} {{lim}}\frac{{n}.\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}\right)}{\left({n}−\frac{\mathrm{1}}{\pi}\right)\left({n}+\mathrm{1}−\frac{\mathrm{1}}{\pi}\right)…\left(\mathrm{2}{n}−\frac{\mathrm{1}}{\pi}\right)} \\ $$$$\:\:\:=\underset{{n}\rightarrow+\infty} {{lim}}\frac{\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\pi}\right)}{\Gamma\left(\mathrm{2}{n}−\frac{\mathrm{1}}{\pi}+\mathrm{1}\right)\Gamma\left({n}\right)}=\underset{{n}\rightarrow+\infty} {{lim}}\frac{\left({n}\right)^{\mathrm{2}{n}} \left({n}−\frac{\mathrm{1}}{\pi}−\mathrm{1}\right)^{{n}} \mathrm{4}^{\frac{\mathrm{1}}{\pi}} }{\left({n}−\mathrm{1}\right)^{{n}} \left({n}−\frac{\mathrm{1}}{\mathrm{2}\pi}\right)^{\mathrm{2}{n}} } \\ $$$$\:\:=\underset{{n}\rightarrow+\infty} {{lim}}\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{\pi\left({n}−\mathrm{1}\right)}\right)^{{n}−\mathrm{1}} \mathrm{4}^{\frac{\mathrm{1}}{\pi}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\pi{n}}\right)^{\mathrm{2}{n}} }=\sqrt[{\pi}]{\mathrm{4}}\: \\ $$$$ \\ $$

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