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Question Number 32938 by abdo imad last updated on 06/Apr/18
let 0<θ<π find Σ_(n=1) ^∞ ((cos(nθ))/n) 2) find Σ_(n=1) ^∞ ((sin(nθ))/n)
$${let}\:\mathrm{0}<\theta<\pi\:\:{find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({n}\theta\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({n}\theta\right)}{{n}} \\ $$
Commented by abdo imad last updated on 09/Apr/18
let put w(z)= Σ_(n=0) ^∞ z^n with ∣z∣≤1 and ∈C−R we have w(z) = (1/(1−z)) ⇒ ∫w(z)dz= −ln (1−z) ⇒ Σ_(n=0) ^∞ (z^(n+1) /(n+1)) = −ln(1−z) ⇒ Σ_(n=1) ^∞ (z^n /n) =−ln(1−z) and for z=e^(iθ) we get Σ_(n=1) ^∞ (e^(iθ) /n) =−ln(1−e^(iθ) ) but ln(1−e^(iθ) ) =ln( 1 −cosθ −i sinθ) = ln( 2 sin^2 ((θ/2)) −2i cos((θ/2))sin((θ/2))) =ln( −2i sin((θ/2)) e^(i(θ/2)) ) =i(θ/2) + ln(−2i) +ln(sin((θ/2))) = ln(2) +ln(sin((θ/2))) +i (θ/2) +ln(e^(−i(π/2)) ) =ln(2 sin((θ/2))) +i ((θ−π)/2) ⇒Σ_(n=1) ^∞ (e^(inθ) /n) =−ln(2sin((θ/2))) +i ((π−θ)/2) ⇒ Σ_(n=1) ^∞ ((cos(nθ))/n) =−ln(2sin((θ/2))) and Σ_(n=1) ^∞ ((sin(nθ))/n) = ((π−θ)/2) .
$${let}\:{put}\:\:{w}\left({z}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{{n}} \:\:\:{with}\:\mid{z}\mid\leqslant\mathrm{1}\:{and}\:\in{C}−{R}\:{we}\:{have} \\ $$$${w}\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}}\:\:\:\Rightarrow\:\int{w}\left({z}\right){dz}=\:−{ln}\:\left(\mathrm{1}−{z}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{z}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:−{ln}\left(\mathrm{1}−{z}\right)\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{z}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{z}\right) \\ $$$${and}\:{for}\:{z}={e}^{{i}\theta} \:{we}\:{get}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{{i}\theta} }{{n}}\:=−{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right) \\ $$$${but}\:\:{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right)\:={ln}\left(\:\mathrm{1}\:−{cos}\theta\:−{i}\:{sin}\theta\right) \\ $$$$=\:{ln}\left(\:\mathrm{2}\:{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\:−\mathrm{2}{i}\:{cos}\left(\frac{\theta}{\mathrm{2}}\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$$$={ln}\left(\:−\mathrm{2}{i}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \right)\:={i}\frac{\theta}{\mathrm{2}}\:+\:{ln}\left(−\mathrm{2}{i}\right)\:+{ln}\left({sin}\left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$$$=\:{ln}\left(\mathrm{2}\right)\:+{ln}\left({sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:+{i}\:\frac{\theta}{\mathrm{2}}\:\:+{ln}\left({e}^{−{i}\frac{\pi}{\mathrm{2}}} \right) \\ $$$$={ln}\left(\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:+{i}\:\frac{\theta−\pi}{\mathrm{2}}\:\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{{in}\theta} }{{n}} \\ $$$$=−{ln}\left(\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:+{i}\:\frac{\pi−\theta}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({n}\theta\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{sin}\left({n}\theta\right)}{{n}}\:=\:\frac{\pi−\theta}{\mathrm{2}}\:\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 09/Apr/18
Σ_(n=1) ^∞ (e^(inθ) /n) = −ln(1−e^(iθ) ) .
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{e}^{{in}\theta} }{{n}}\:=\:−{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right)\:. \\ $$

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