Question Number 67232 by prof Abdo imad last updated on 24/Aug/19
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({n}\frac{\pi}{\mathrm{3}}\right)}{{n}} \\ $$
Commented by mathmax by abdo last updated on 25/Aug/19
$${first}\:{let}\:{determine}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}\:={Re}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right) \\ $$$${let}\:{s}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\:\Rightarrow{s}^{'} \left({x}\right)={i}\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{inx}} ={ie}^{{ix}} \sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{i}\left({n}−\mathrm{1}\right){x}} \\ $$$$={ie}^{{ix}} \sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{{inx}} \:=\frac{{ie}^{{ix}} }{\mathrm{1}−{e}^{{ix}} }\:\:\Rightarrow{s}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right)+{c} \\ $$$${x}=\pi\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right)\:=−{ln}\left(\mathrm{2}\right)+{c}\:\Rightarrow{c}\:=\mathrm{0} \\ $$$${s}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$${ln}\left(\mathrm{1}−{e}^{{ix}} \right)={ln}\left(\mathrm{1}−{cosx}−{isinx}\right)\:={ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$={ln}\left(−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){e}^{{i}\frac{{x}}{\mathrm{2}}} \right)\:={ln}\left(−\mathrm{2}{i}\right)+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)+{i}\frac{{x}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}\right)+{ln}\left(−{i}\right)\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)+\frac{{ix}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{i}\pi}{\mathrm{2}}\:+\frac{{ix}}{\mathrm{2}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$${x}=\frac{\pi}{\mathrm{3}}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)\:=−{ln}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{0}\:{so} \\ $$$${we}\:{have}\:\:\frac{{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}}\:+\frac{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}{\mathrm{2}}\:+\frac{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{3}}\right)}{\mathrm{3}}\:+….=\mathrm{0} \\ $$$${also}\:{we}\:{can}\:{extract}\:{that}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}\:=\frac{\pi−\frac{\pi}{\mathrm{3}}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\:=\frac{\mathrm{2}\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$
Answered by Smail last updated on 25/Aug/19
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{inx}} }{{n}}=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{cos}\left({x}\right)−{isin}\left({x}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left({x}/\mathrm{2}\right)−\mathrm{2}{isin}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{2}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid−{ln}\left({sin}\left({x}/\mathrm{2}\right)−{icos}\left({x}/\mathrm{2}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid−{ln}\left({cos}\left(\frac{\pi−{x}}{\mathrm{2}}\right)−{isin}\left(\frac{\pi−{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid+\frac{\pi−{x}}{\mathrm{2}}{i} \\ $$
Commented by Smail last updated on 25/Aug/19
$${For}\:{x}=\frac{\pi}{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\frac{\pi}{\mathrm{3}}} }{{n}}=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left(\pi/\mathrm{6}\right)\mid+{i}\frac{\pi−\pi/\mathrm{3}}{\mathrm{2}} \\ $$$${So},\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({n}\frac{\pi}{\mathrm{3}}\right)}{{n}}={Re}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\frac{\pi}{\mathrm{3}}} }{{n}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left(\pi/\mathrm{6}\right)\mid \\ $$