Question Number 117458 by Dwaipayan Shikari last updated on 11/Oct/20
$$\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{16}}{\mathrm{15}}.\frac{\mathrm{36}}{\mathrm{35}}.\frac{\mathrm{64}}{\mathrm{63}}.\frac{\mathrm{100}}{\mathrm{99}}.\frac{\mathrm{144}}{\mathrm{143}}.\frac{\mathrm{196}}{\mathrm{195}}.\frac{\mathrm{256}}{\mathrm{255}}.\frac{\mathrm{324}}{\mathrm{323}}……\infty \\ $$
Answered by Olaf last updated on 11/Oct/20
![P_n = Π_(k=1) ^n (((2k)^2 )/((2k−1)(2k+1))) P_n = Π_(k=1) ^n (((2k)^4 )/([(2k−1)(2k)][(2k)(2k+1)])) P_n = ((2^(4n) n!^4 )/((2n)!(2n+1)!)) = ((2^(4n) n!^4 )/((2n)!^2 (2n+1))) n! ∼ (√(2πn))((n/e))^n (Stirling) P_n ∼ ((2^(4n) 4π^2 n^2 ((n/e))^(4n) )/( 2π(2n)[(((2n)/e))^(2n) ]^2 (2n+1))) P_n ∼ ((πn)/( 2n+1)) ∼ (π/2) lim_(n→∞) P_n = (π/2)](https://www.tinkutara.com/question/Q117473.png)
$$\mathrm{P}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\left(\mathrm{2}{k}\right)^{\mathrm{2}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$\mathrm{P}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\left(\mathrm{2}{k}\right)^{\mathrm{4}} }{\left[\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}\right)\right]\left[\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\right]} \\ $$$$\mathrm{P}_{{n}} \:=\:\frac{\mathrm{2}^{\mathrm{4}{n}} {n}!^{\mathrm{4}} }{\left(\mathrm{2}{n}\right)!\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:=\:\frac{\mathrm{2}^{\mathrm{4}{n}} {n}!^{\mathrm{4}} }{\left(\mathrm{2}{n}\right)!^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$${n}!\:\sim\:\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} \:\left(\mathrm{Stirling}\right) \\ $$$$\mathrm{P}_{{n}} \:\sim\:\frac{\mathrm{2}^{\mathrm{4}{n}} \mathrm{4}\pi^{\mathrm{2}} {n}^{\mathrm{2}} \left(\frac{{n}}{{e}}\right)^{\mathrm{4}{n}} }{\:\mathrm{2}\pi\left(\mathrm{2}{n}\right)\left[\left(\frac{\mathrm{2}{n}}{{e}}\right)^{\mathrm{2}{n}} \right]^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\mathrm{P}_{{n}} \:\sim\:\frac{\pi{n}}{\:\mathrm{2}{n}+\mathrm{1}}\:\sim\:\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}P}_{{n}} \:=\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 11/Oct/20
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{4}{n}^{\mathrm{2}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}={y} \\ $$$$\frac{\mathrm{1}}{{y}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\:\:\:\:\:\left({z}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${sinz}\pi=\pi{z}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${sin}\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\right) \\ $$$$\mathrm{1}=\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{{y}} \\ $$$${y}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$