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Question Number 38109 by maxmathsup by imad last updated on 21/Jun/18

1) find S(x) = Σ_(n=1) ^∞ ((cos(nx))/n) 2) find Σ_(n=1) ^∞ (((−1)^n )/n)

$$\left.\mathrm{1}\right)\:{find}\:\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$ \\ $$

Commented by abdo.msup.com last updated on 30/Jun/18

we have S(x)=Re(Σ_(n=1) ^∞ (e^(inx) /n))=Re(w(x)) w(x)=Σ_(n=1) ^∞ (e^(inx) /n) ⇒w^′ (x)=i Σ_(n=1) ^∞ (e^(ix) )^n =i e^(ix) Σ_(n=1) ^∞ (e^(ix) )^(n−1) =ie^(ix) Σ_(n=0) ^∞ (e^(ix) )^n =i(e^(ix) /(1−e^(ix) )) =i (e^(ix) /(1−cosx −isinx)) =i (e^(ix) /(2sin^2 ((x/2))−2isin((x/2))cos((x/2)))) = (i/(−2isin((x/2)))) (e^(ix) /e^(i(x/2)) ) =((−1)/(2 sin((x/2)))) e^(i(x/2)) = ((−1)/(2sin((x/2)))){ cos((x/2)) +i sin((x/2))} =−(1/2)cotan((x/2)) −(i/2) ⇒ w(x)=−(1/2) ∫ ((cos((x/2)))/(sin((x/2)))) −((ix)/2) +c =−ln∣sin((x/2))∣ −((ix)/2) +c w(π) =−((iπ)/2) +c =Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) ⇒c=((iπ)/2) −ln(2) ⇒ w(x)=−ln∣ sin((x/2))∣−((ix)/2) +((iπ)/2) −ln(2) S(x)=Re(w(x))=−ln∣sin((x/2))∣−ln(2). also we get Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) .

$${we}\:{have}\:{S}\left({x}\right)={Re}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\right)={Re}\left({w}\left({x}\right)\right) \\ $$$${w}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}}\:\Rightarrow{w}^{'} \left({x}\right)={i}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}} \\ $$$$={i}\:{e}^{{ix}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}−\mathrm{1}} ={ie}^{{ix}} \sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{{ix}} \right)^{{n}} \\ $$$$={i}\frac{{e}^{{ix}} }{\mathrm{1}−{e}^{{ix}} }\:={i}\:\frac{{e}^{{ix}} }{\mathrm{1}−{cosx}\:−{isinx}} \\ $$$$={i}\:\frac{{e}^{{ix}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=\:\frac{{i}}{−\mathrm{2}{isin}\left(\frac{{x}}{\mathrm{2}}\right)}\:\frac{{e}^{{ix}} }{{e}^{{i}\frac{{x}}{\mathrm{2}}} } \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:{e}^{{i}\frac{{x}}{\mathrm{2}}} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\left\{\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{cotan}\left(\frac{{x}}{\mathrm{2}}\right)\:−\frac{{i}}{\mathrm{2}}\:\Rightarrow \\ $$$${w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\:−\frac{{ix}}{\mathrm{2}}\:+{c} \\ $$$$=−{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:−\frac{{ix}}{\mathrm{2}}\:+{c}\: \\ $$$${w}\left(\pi\right)\:=−\frac{{i}\pi}{\mathrm{2}}\:+{c}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow{c}=\frac{{i}\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${w}\left({x}\right)=−{ln}\mid\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid−\frac{{ix}}{\mathrm{2}}\:+\frac{{i}\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right) \\ $$$${S}\left({x}\right)={Re}\left({w}\left({x}\right)\right)=−{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid−{ln}\left(\mathrm{2}\right). \\ $$$${also}\:{we}\:{get}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi−{x}}{\mathrm{2}}\:. \\ $$

Commented by abdo.msup.com last updated on 30/Jun/18

2) we have provef that Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n =ln(1+x) if ∣x∣≤1 and x≠−1 ⇒Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(2) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2).

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{provef}\:{that} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \:={ln}\left(\mathrm{1}+{x}\right)\:{if}\:\mid{x}\mid\leqslant\mathrm{1}\:{and} \\ $$$${x}\neq−\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:={ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right). \\ $$

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