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Question-160496

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Question Number 160496 by mathlove last updated on 30/Nov/21
Answered by Kamel last updated on 01/Dec/21
z^(2n+1) −1=Π_(k=0) ^(2n) (z−e^((2kπi)/(2n+1)) ) ∴ Σ_(k=0) ^(2n) z^k =Π_(k=1) ^(2n) (z−e^((2kπi)/(2n+1)) ) So: for z=1, 2n+1=Π_(k=1) ^(2n) (e^((2kπi)/(2n+1)) −1) For z=−1, 1=Π_(k=1) ^(2n) (e^((2kπi)/(2n+1)) +1) tan(((kπ)/(2n+1)))=((e^((2kπi)/(2n+1)) −1)/(i(e^((2kπi)/(2n+1)) +1))) ∴ Π_(k=1) ^(2n) (−1)^n tan(((kπ)/(2n+1)))=2n+1=^(p=2n+1−k) (−1)^n Π_(k=1) ^n tan(((kπ)/(2n+1)))Π_(p=1) ^n tan(π−((pπ)/(2n+1))) Then: 2n+1=(Π_(k=1) ^n tan(((kπ)/(2n+1))))^2 Or: tan((π/(2n+1)))tan(((2π)/(2n+1)))...tan(((nπ)/(2n+1)))=(√(2n+1))
$$ \\ $$$${z}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\prod}}\left({z}−{e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} \right) \\ $$$$\therefore\:\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{z}^{{k}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\left({z}−{e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} \right) \\ $$$${So}:\:{for}\:{z}=\mathrm{1},\:\mathrm{2}{n}+\mathrm{1}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\left({e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} −\mathrm{1}\right) \\ $$$${For}\:{z}=−\mathrm{1},\:\mathrm{1}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\left({e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} +\mathrm{1}\right) \\ $$$${tan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\frac{{e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} −\mathrm{1}}{{i}\left({e}^{\frac{\mathrm{2}{k}\pi{i}}{\mathrm{2}{n}+\mathrm{1}}} +\mathrm{1}\right)} \\ $$$$\therefore\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\prod}}\left(−\mathrm{1}\right)^{{n}} {tan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\mathrm{2}{n}+\mathrm{1}\overset{{p}=\mathrm{2}{n}+\mathrm{1}−{k}} {=}\left(−\mathrm{1}\right)^{{n}} \underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}{tan}\left(\pi−\frac{{p}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$$${Then}:\:\mathrm{2}{n}+\mathrm{1}=\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\right)^{\mathrm{2}} \\ $$$${Or}:\:{tan}\left(\frac{\pi}{\mathrm{2}{n}+\mathrm{1}}\right){tan}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)…{tan}\left(\frac{{n}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{n}+\mathrm{1}} \\ $$
Answered by mr W last updated on 30/Nov/21
(2) lim_(n→∞) Σ_(k=1) ^n (1/(n+k)) =lim_(n→∞) (1/n)Σ_(k=1) ^n (1/((k/n)+1)) =∫_0 ^1 (dx/(x+1)) =ln (x+1)∣_0 ^1 =ln 2
$$\left(\mathrm{2}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}+{k}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\frac{{k}}{{n}}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{{x}+\mathrm{1}} \\ $$$$=\mathrm{ln}\:\left({x}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\:\mathrm{2} \\ $$

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