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Question Number 67232 by prof Abdo imad last updated on 24/Aug/19

calculate Σ_(n=1) ^∞ ((cos(n(π/3)))/n)

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({n}\frac{\pi}{\mathrm{3}}\right)}{{n}} \\ $$

Commented by mathmax by abdo last updated on 25/Aug/19

first let determine Σ_(n=1) ^∞ ((cos(nx))/n) =Re(Σ_(n=1) ^∞ (e^(inx) /n)) let s(x) =Σ_(n=1) ^∞ (e^(inx) /n) ⇒s^′ (x)=iΣ_(n=1) ^∞ e^(inx) =ie^(ix) Σ_(n=1) ^∞ e^(i(n−1)x) =ie^(ix) Σ_(n=0) ^∞ e^(inx) =((ie^(ix) )/(1−e^(ix) )) ⇒s(x) =−ln(1−e^(ix) )+c x=π ⇒Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) =−ln(2)+c ⇒c =0 s(x) =−ln(1−e^(ix) ) ln(1−e^(ix) )=ln(1−cosx−isinx) =ln(2sin^2 ((x/2))−2isin((x/2))cos((x/2))) =ln(−2isin((x/2))e^(i(x/2)) ) =ln(−2i)+ln(sin((x/2)))+i(x/2) =ln(2)+ln(−i) +ln(sin((x/2)))+((ix)/2) =ln(2sin((x/2)))−((iπ)/2) +((ix)/2) ⇒Σ_(n=1) ^∞ ((cos(nx))/n) =−ln(2sin((x/2))) x=(π/3) ⇒ Σ_(n=1) ^∞ ((cos(((nπ)/3)))/n) =−ln(2sin((π/6))) =−ln(2×(1/2)) =0 so we have ((cos((π/3)))/1) +((cos(((2π)/3)))/2) +((cos(((3π)/3)))/3) +....=0 also we can extract that Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) ⇒ Σ_(n=1) ^∞ ((sin(((nπ)/3)))/n) =((π−(π/3))/2) =(π/2)−(π/6) =((2π)/6) =(π/3)

$${first}\:{let}\:{determine}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}\:={Re}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right) \\ $$$${let}\:{s}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\:\Rightarrow{s}^{'} \left({x}\right)={i}\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{inx}} ={ie}^{{ix}} \sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{i}\left({n}−\mathrm{1}\right){x}} \\ $$$$={ie}^{{ix}} \sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{{inx}} \:=\frac{{ie}^{{ix}} }{\mathrm{1}−{e}^{{ix}} }\:\:\Rightarrow{s}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right)+{c} \\ $$$${x}=\pi\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right)\:=−{ln}\left(\mathrm{2}\right)+{c}\:\Rightarrow{c}\:=\mathrm{0} \\ $$$${s}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$${ln}\left(\mathrm{1}−{e}^{{ix}} \right)={ln}\left(\mathrm{1}−{cosx}−{isinx}\right)\:={ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$={ln}\left(−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){e}^{{i}\frac{{x}}{\mathrm{2}}} \right)\:={ln}\left(−\mathrm{2}{i}\right)+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)+{i}\frac{{x}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}\right)+{ln}\left(−{i}\right)\:+{ln}\left({sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)+\frac{{ix}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{{i}\pi}{\mathrm{2}}\:+\frac{{ix}}{\mathrm{2}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({nx}\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$${x}=\frac{\pi}{\mathrm{3}}\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}\:=−{ln}\left(\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)\:=−{ln}\left(\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\mathrm{0}\:{so} \\ $$$${we}\:{have}\:\:\frac{{cos}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}}\:+\frac{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}{\mathrm{2}}\:+\frac{{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{3}}\right)}{\mathrm{3}}\:+....=\mathrm{0} \\ $$$${also}\:{we}\:{can}\:{extract}\:{that}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}\:=\frac{\pi−\frac{\pi}{\mathrm{3}}}{\mathrm{2}}\:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\:=\frac{\mathrm{2}\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$

Answered by Smail last updated on 25/Aug/19

Σ_(n=1) ^∞ (e^(inx) /n)=−ln(1−e^(ix) ) =−ln(1−cos(x)−isin(x)) =−ln(2sin^2 (x/2)−2isin(x/2)cos(x/2)) =−ln(2)−ln∣sin(x/2)∣−ln(sin(x/2)−icos(x/2)) =−ln(2)−ln∣sin(x/2)∣−ln(cos(((π−x)/2))−isin(((π−x)/2))) =−ln(2)−ln∣sin(x/2)∣+((π−x)/2)i

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{inx}} }{{n}}=−{ln}\left(\mathrm{1}−{e}^{{ix}} \right) \\ $$$$=−{ln}\left(\mathrm{1}−{cos}\left({x}\right)−{isin}\left({x}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}{sin}^{\mathrm{2}} \left({x}/\mathrm{2}\right)−\mathrm{2}{isin}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{2}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid−{ln}\left({sin}\left({x}/\mathrm{2}\right)−{icos}\left({x}/\mathrm{2}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid−{ln}\left({cos}\left(\frac{\pi−{x}}{\mathrm{2}}\right)−{isin}\left(\frac{\pi−{x}}{\mathrm{2}}\right)\right) \\ $$$$=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left({x}/\mathrm{2}\right)\mid+\frac{\pi−{x}}{\mathrm{2}}{i} \\ $$

Commented by Smail last updated on 25/Aug/19

For x=(π/3) Σ_(n=1) ^∞ (e^(in(π/3)) /n)=−ln(2)−ln∣sin(π/6)∣+i((π−π/3)/2) So, Σ_(n=1) ^∞ ((cos(n(π/3)))/n)=Re(Σ_(n=1) ^∞ (e^(in(π/3)) /n)) Σ_(n=1) ^∞ ((cos(((nπ)/3)))/n)=−ln(2)−ln∣sin(π/6)∣

$${For}\:{x}=\frac{\pi}{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\frac{\pi}{\mathrm{3}}} }{{n}}=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left(\pi/\mathrm{6}\right)\mid+{i}\frac{\pi−\pi/\mathrm{3}}{\mathrm{2}} \\ $$$${So},\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({n}\frac{\pi}{\mathrm{3}}\right)}{{n}}={Re}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\frac{\pi}{\mathrm{3}}} }{{n}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{{n}}=−{ln}\left(\mathrm{2}\right)−{ln}\mid{sin}\left(\pi/\mathrm{6}\right)\mid \\ $$

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