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Question Number 32938 by abdo imad last updated on 06/Apr/18 | ||
$${let}\:\mathrm{0}<\theta<\pi\:\:{find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({n}\theta\right)}{{n}} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({n}\theta\right)}{{n}} \\ $$ | ||
Commented byabdo imad last updated on 09/Apr/18 | ||
$${let}\:{put}\:\:{w}\left({z}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{z}^{{n}} \:\:\:{with}\:\mid{z}\mid\leqslant\mathrm{1}\:{and}\:\in{C}−{R}\:{we}\:{have} \\ $$ $${w}\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{z}}\:\:\:\Rightarrow\:\int{w}\left({z}\right){dz}=\:−{ln}\:\left(\mathrm{1}−{z}\right)\:\Rightarrow \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{z}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=\:−{ln}\left(\mathrm{1}−{z}\right)\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{z}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{z}\right) \\ $$ $${and}\:{for}\:{z}={e}^{{i}\theta} \:{we}\:{get}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{{i}\theta} }{{n}}\:=−{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right) \\ $$ $${but}\:\:{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right)\:={ln}\left(\:\mathrm{1}\:−{cos}\theta\:−{i}\:{sin}\theta\right) \\ $$ $$=\:{ln}\left(\:\mathrm{2}\:{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\:−\mathrm{2}{i}\:{cos}\left(\frac{\theta}{\mathrm{2}}\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$ $$={ln}\left(\:−\mathrm{2}{i}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \right)\:={i}\frac{\theta}{\mathrm{2}}\:+\:{ln}\left(−\mathrm{2}{i}\right)\:+{ln}\left({sin}\left(\frac{\theta}{\mathrm{2}}\right)\right) \\ $$ $$=\:{ln}\left(\mathrm{2}\right)\:+{ln}\left({sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:+{i}\:\frac{\theta}{\mathrm{2}}\:\:+{ln}\left({e}^{−{i}\frac{\pi}{\mathrm{2}}} \right) \\ $$ $$={ln}\left(\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:+{i}\:\frac{\theta−\pi}{\mathrm{2}}\:\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{{in}\theta} }{{n}} \\ $$ $$=−{ln}\left(\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:+{i}\:\frac{\pi−\theta}{\mathrm{2}}\:\Rightarrow \\ $$ $$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({n}\theta\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:{and} \\ $$ $$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{sin}\left({n}\theta\right)}{{n}}\:=\:\frac{\pi−\theta}{\mathrm{2}}\:\:. \\ $$ $$ \\ $$ | ||
Commented byprof Abdo imad last updated on 09/Apr/18 | ||
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{e}^{{in}\theta} }{{n}}\:=\:−{ln}\left(\mathrm{1}−{e}^{{i}\theta} \right)\:. \\ $$ | ||