# lim-n-n-0-pi-2-cos-2n-1-d-

Question Number 131548 by liberty last updated on 06/Feb/21
$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{n}}\:\underset{\mathrm{0}} {\overset{\:\:\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:^{\mathrm{2n}+\mathrm{1}} \left(\theta\right)\:\mathrm{d}\theta\:=? \\$$
Answered by rs4089 last updated on 06/Feb/21
$$\frac{\pi\sqrt{\pi}}{\mathrm{2}} \\$$
Commented by liberty last updated on 06/Feb/21
$$\mathrm{False} \\$$
Answered by liberty last updated on 06/Feb/21
$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\theta\right)\:\mathrm{cos}\:^{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{1}} \left(\theta\right)\mathrm{d}\theta \\$$$$\:\:\:=\:\:\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\$$$$\mathrm{L}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{n}\pi}}{\mathrm{2}}.\:\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right)}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\$$$$\mathrm{L}=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{n}\pi}}{\mathrm{2}}\:\sqrt{\frac{\mathrm{e}}{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}}}\:\sqrt{\frac{\mathrm{2n}}{\mathrm{2n}+\mathrm{1}}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}\right)^{−\mathrm{n}} \\$$$$\mathrm{L}=\:\frac{\sqrt{\pi\mathrm{e}}}{\mathrm{2}}.\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}\right)^{−\mathrm{n}} \\$$$$\mathrm{L}\:=\:\frac{\sqrt{\pi\mathrm{e}}}{\mathrm{2}}\:.\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}} \\$$$$\\$$
Commented by rs4089 last updated on 06/Feb/21
$${thank}\:{you}\:{very}\:{much}\:{sir}\: \\$$$$\left.{i}\:{got}\:{my}\:{mistake}:\right) \\$$
Answered by rs4089 last updated on 06/Feb/21
$${lim}_{{n}\rightarrow\infty} \sqrt{{n}\:}\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\$$$${lim}_{{n}\rightarrow\infty} \sqrt{{n}}\frac{\:{n}!\:\:\sqrt{\pi}}{\mathrm{2}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\$$$${lim}_{{n}\rightarrow\infty} \sqrt{{n}}\frac{{n}!\sqrt{\pi}\left(\mathrm{4}^{{n}} {n}!\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\:\left(\mathrm{2}{n}!\:\sqrt{\pi}\right)} \\$$$${lim}_{{n}\rightarrow\infty} \sqrt{{n}}\frac{\mathrm{4}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}{n}} {C}_{{n}} } \\$$$${lim}_{{n}\rightarrow\infty} \sqrt{{n}}\frac{\mathrm{4}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\frac{\mathrm{4}^{{n}} }{\:\sqrt{\pi{n}}}\right)} \\$$$${lim}_{{n}\rightarrow\infty} \frac{{n}\sqrt{\pi}}{\mathrm{2}{n}+\mathrm{1}}\:\:=\:\frac{\sqrt{\pi}}{\mathrm{2}} \\$$$$\\$$
Answered by mathmax by abdo last updated on 06/Feb/21
$$\mathrm{let}\:\:\mathrm{W}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}+\mathrm{1}} \theta\:\mathrm{d}\theta\:\Rightarrow\mathrm{W}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2n}+\mathrm{1}} \theta\:.\mathrm{sin}^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \theta\mathrm{d}\theta \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{cos}^{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{1}} \theta\:\mathrm{sin}^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \theta\mathrm{d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B}\left(\mathrm{n}+\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\mathrm{n}+\mathrm{1}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\mathrm{n}!}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\mathrm{we}\:\mathrm{have} \\$$$$\mathrm{n}!\:\sim\mathrm{n}^{\mathrm{n}} .\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}\:\:\mathrm{and}\:\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\:=\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)!\sim\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \sqrt{\mathrm{2}\pi\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\: \\$$$$\Rightarrow\frac{\mathrm{n}!}{\Gamma\left(\mathrm{n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\sim\frac{\mathrm{n}^{\mathrm{n}} .\mathrm{e}^{−\mathrm{n}} \sqrt{\mathrm{2}\pi\mathrm{n}}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{n}} .\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{2}\pi}.\sqrt{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}}} \\$$$$=\sqrt{\mathrm{e}}.\frac{\mathrm{n}^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} }\:\Rightarrow\sqrt{\mathrm{n}}\mathrm{W}_{\mathrm{n}} \sim\:\sqrt{\mathrm{e}}.\frac{\mathrm{n}^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}+\mathrm{1}} } \\$$$$=\sqrt{\mathrm{e}}.\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{n}+\mathrm{1}} \:=\sqrt{\mathrm{e}}\mathrm{e}^{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{log}\left(\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)} \:=\sqrt{\mathrm{e}}\mathrm{e}^{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)} \\$$$$=\sqrt{\mathrm{e}}.\mathrm{e}^{−\mathrm{1}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{nW}_{\mathrm{n}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\$$
Commented by mathmax by abdo last updated on 06/Feb/21
$$\mathrm{sorry}\:\:\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{nW}_{\mathrm{n}} =\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{e}}} \\$$