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Question Number 50426 by Abdo msup. last updated on 16/Dec/18
study the convergence of U_n =((2/π) ∫_0 ^(π/2) (sinx)^(1/n) )^n
$${study}\:{the}\:{convergence}\:{of}\: \\ $$$${U}_{{n}} =\left(\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \right)^{{n}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18
(1/π)×2∫_0 ^(π/2) sinx)^(2×(((1+n)/(2n)))−1) (cosx)^(2×(1/2)−1) dx formula 2∫_0 ^(π/2) sin^(2p−1) xcos^(2q−1) xdx =((⌈(p)⌈q))/(⌈p+q))) so (1/π)×((⌈(((1+n)/(2n)))×⌈((1/2)))/(⌈(((1+n)/(2n))+(1/2)))) =(1/( (√π)))×((⌈((1/(2n))+(1/2)))/(⌈((1/(2n))+1))) wait...
$$\left.\frac{\mathrm{1}}{\pi}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sinx}\right)^{\mathrm{2}×\left(\frac{\mathrm{1}+{n}}{\mathrm{2}{n}}\right)−\mathrm{1}} \left({cosx}\right)^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{q}−\mathrm{1}} {xdx} \\ $$$$=\frac{\left.\lceil\left({p}\right)\lceil{q}\right)}{\left.\lceil{p}+{q}\right)} \\ $$$${so}\:\frac{\mathrm{1}}{\pi}×\frac{\lceil\left(\frac{\mathrm{1}+{n}}{\mathrm{2}{n}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{\mathrm{1}+{n}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\pi}}×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}{n}}+\mathrm{1}\right)} \\ $$$${wait}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18
Another approach... 1≥sinx≥0 1≥(sinx)^(1/n) ≥0 (2/π)×∫_0 ^(π/2) 1×dx≥(2/π)∫_0 ^(π/2) (sinx)^(1/n) dx≥(2/π)∫_0 ^(π/2) 0×dx 2≥(2/π)∫_0 ^(π/2) (sinx)^(1/n) dx≥0 2^n ≥[(2/π)∫_0 ^(π/2) (sinx)^(1/n) dx]^n ≥0
$${Another}\:{approach}… \\ $$$$\:\mathrm{1}\geqslant{sinx}\geqslant\mathrm{0} \\ $$$$\mathrm{1}\geqslant\left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \geqslant\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\pi}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{1}×{dx}\geqslant\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} {dx}\geqslant\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{0}×{dx} \\ $$$$\mathrm{2}\geqslant\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} {dx}\geqslant\mathrm{0} \\ $$$$\mathrm{2}^{{n}} \geqslant\left[\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} {dx}\right]^{{n}} \geqslant\mathrm{0} \\ $$

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