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Question Number 117086 by mnjuly1970 last updated on 09/Oct/20

...nice mathematics... evaluate ... lim_(n→∞) (Π_(k=1) ^n sin(((kπ)/(4n))))^(1/n) =? ...m.n.1970...

$$\:\:\:\:\:\:\:...{nice}\:\:{mathematics}... \\ $$$$\:{evaluate}\:... \\ $$$$\:\:\:\:\:\:\:{lim}_{{n}\rightarrow\infty} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{4}{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 09/Oct/20

Commented by mnjuly1970 last updated on 09/Oct/20

thank you very much..

$${thank}\:{you}\:{very}\:{much}.. \\ $$

Commented by Ar Brandon last updated on 09/Oct/20

You're welcome Sir ��

Answered by Dwaipayan Shikari last updated on 09/Oct/20

lim_(n→∞) (Π_(k=1) ^n sin(((kπ)/(4n))))^(1/n) =y lim_(n→∞) (1/n)Σ_(k=1) ^n log(sin(((kπ)/(4n))))=logy ∫_0 ^1 log(sin(((πx)/4)))=logy (4/π)∫_0 ^(π/4) log(sint)dt −1.27=logy (Wolfram) y=e^(−1.27)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{sin}\left(\frac{{k}\pi}{\mathrm{4}{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} ={y} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{log}\left({sin}\left(\frac{{k}\pi}{\mathrm{4}{n}}\right)\right)={logy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left({sin}\left(\frac{\pi{x}}{\mathrm{4}}\right)\right)={logy} \\ $$$$\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {log}\left({sint}\right){dt} \\ $$$$−\mathrm{1}.\mathrm{27}={logy}\:\:\left({Wolfram}\right) \\ $$$${y}={e}^{−\mathrm{1}.\mathrm{27}} \\ $$

Commented by Olaf last updated on 10/Oct/20

∫_0 ^(π/4) ln(sint)dt = −(K/2)−(π/4)ln2 exactly. (with K = β(2) : constant of Catalan) So limit is (1/2)e^(−((2β(2))/π))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sin}{t}\right){dt}\:=\:−\frac{{K}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}\:\mathrm{exactly}. \\ $$$$\left(\mathrm{with}\:{K}\:=\:\beta\left(\mathrm{2}\right)\::\:\mathrm{constant}\:\mathrm{of}\:\mathrm{Catalan}\right) \\ $$$$\mathrm{So}\:\mathrm{limit}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\mathrm{2}\beta\left(\mathrm{2}\right)}{\pi}} \\ $$

Commented by mnjuly1970 last updated on 09/Oct/20

thank you so much. answer :=e^((((−2G)/π)−log(2))) ✓ recall: ∫_0 ^( (π/4)) log(sin(x))dx =((−G)/2) −(π/4) log(2) m.n.1970 sincerely yours...

$${thank}\:{you}\:{so}\:{much}. \\ $$$$\:{answer}\::={e}^{\left(\frac{−\mathrm{2}{G}}{\pi}−{log}\left(\mathrm{2}\right)\right)} \:\checkmark \\ $$$${recall}:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {log}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−{G}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:{log}\left(\mathrm{2}\right) \\ $$$$\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$\:\:{sincerely}\:\:{yours}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by mnjuly1970 last updated on 10/Oct/20

you are right. your solution is simplified...

$${you}\:{are}\:{right}. \\ $$$${your}\:{solution}\:{is} \\ $$$${simplified}... \\ $$

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