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Question Number 33131 by prof Abdo imad last updated on 11/Apr/18

1)find  Σ_(n=1) ^∞   (e^(inx) /(n(n+1)))  2) find the value of   Σ_(n≥1)  ((sin(nx))/(n(n+1)))  and Σ_(n≥1)   ((cos(nx))/(n(n+1))) .

$$\left.\mathrm{1}\right){find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{{inx}} }{{n}\left({n}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{{sin}\left({nx}\right)}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${and}\:\sum_{{n}\geqslant\mathrm{1}} \:\:\frac{{cos}\left({nx}\right)}{{n}\left({n}+\mathrm{1}\right)}\:. \\ $$

Commented by prof Abdo imad last updated on 12/Apr/18

let find S(x) = Σ_(n=1) ^∞   (x^n /(n(n+1)))  the radius of   convergrnce is 1 and  for x=+^− 1 the serie is  also convergent we have for  ∣x∣<1  S(x)= Σ_(n=1) ^∞   ((1/n) −(1/(n+1)))x^n  =Σ_(n=1) ^∞  (x^n /n) −Σ_(n=1) ^∞   (x^n /(n+1))  but  Σ_(n=1) ^∞   (x^n /n) =−ln(1−x)  Σ_(n=1) ^∞    (x^n /(n+1)) = Σ_(n=2) ^∞   (x^(n−1) /n) = (1/x) Σ_(n=2) ^∞   (x^n /n)  =(1/x)( −ln(1−x) −x)  =−(1/x)ln(1−x) −1 ⇒  S(x)= −ln(1−x) +(1/x)ln(1−x) +1  =(−1 +(1/x))ln(1−x) +1 ⇒  S(x) = ((1−x)/x) ln(1−x) +1  let change x per e^(ix)   we get  Σ_(n=1) ^∞     (e^(inx) /(n(n+1)))  = ((1−e^(ix) )/e^(ix) ) ln(1−e^(ix) ) +1  =( e^(−ix)  −1)ln(1−e^(ix) ) +1  2) Σ_(n=1) ^∞   ((sin(nx))/(n(n+1)))  =Im(S(e^(ix) ))  let find it  (e^(−ix)  −1) = cosx −i sinx −1    = −2sin^2 ((x/2)) −2i sin((x/2))cos((x/2))  = −2i sin((x/2))( cos((x/2)) −i sin((x/2)))  =−2i e^(−i(x/2))   ln(1−e^(ix) ) = ln (1−cosx −isinx)  = ln( 2sin^2 ((x/2)) −2i sin((x/2))cos((x/2)))  =ln(−2i sin((x/2))(cos((x/2)) +isin((x/2)))  = ln(−i) +ln(2sin((x/2)) +ln(e^(i(x/2)) )  =−i (π/2) +ln(2sin((x/2))) +((ix)/2)  =ln(2sin((x/2))) +i((x−π)/2)  Im( S(e^(ix) )) = −2i e^(−i(x/2)) (  ln(2sin((x/2)) +i((x−π)/2))  =−2i( cos((x/2)) −isin((x/2)))(ln(2sin((x/2)) +i((x−π)/2))  = −2i ) ( cos((x/2))ln(2sin((x/2)) +i((x−π)/2) cos((x/2))  −i sin((x/2))ln(2sin((x/2)) +((x−π)/2) sin((x/2)))  = −2i cos((x/2))ln(2sin((x/2)) +(x−π)cos((x/2))  −2 sin((x/2))ln(2sin((x/2)) +i(π−x) sin((x/2)) ⇒  Σ_(n=1) ^∞  ((sin(nx))/(n(n+1)))  =(π−x)sin((x/2)) −sin((x/2))ln(2sin((x/2)))  Σ_(n=1) ^∞   ((cos(nx))/(n(n+1))) =(x−π)cos((x/2)) −2sin((x/2))ln(2sin((x/2))) +1.

$${let}\:{find}\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:\:{the}\:{radius}\:{of}\: \\ $$$${convergrnce}\:{is}\:\mathrm{1}\:{and}\:\:{for}\:{x}=\overset{−} {+}\mathrm{1}\:{the}\:{serie}\:{is} \\ $$$${also}\:{convergent}\:{we}\:{have}\:{for}\:\:\mid{x}\mid<\mathrm{1} \\ $$$${S}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(\frac{\mathrm{1}}{{n}}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}+\mathrm{1}} \\ $$$${but}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}}\:=−{ln}\left(\mathrm{1}−{x}\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:=\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}−\mathrm{1}} }{{n}}\:=\:\frac{\mathrm{1}}{{x}}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}} \\ $$$$=\frac{\mathrm{1}}{{x}}\left(\:−{ln}\left(\mathrm{1}−{x}\right)\:−{x}\right)\:\:=−\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}−{x}\right)\:−\mathrm{1}\:\Rightarrow \\ $$$${S}\left({x}\right)=\:−{ln}\left(\mathrm{1}−{x}\right)\:+\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1} \\ $$$$=\left(−\mathrm{1}\:+\frac{\mathrm{1}}{{x}}\right){ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1}\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\:\frac{\mathrm{1}−{x}}{{x}}\:{ln}\left(\mathrm{1}−{x}\right)\:+\mathrm{1}\:\:{let}\:{change}\:{x}\:{per}\:{e}^{{ix}} \\ $$$${we}\:{get}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{{e}^{{inx}} }{{n}\left({n}+\mathrm{1}\right)}\:\:=\:\frac{\mathrm{1}−{e}^{{ix}} }{{e}^{{ix}} }\:{ln}\left(\mathrm{1}−{e}^{{ix}} \right)\:+\mathrm{1} \\ $$$$=\left(\:{e}^{−{ix}} \:−\mathrm{1}\right){ln}\left(\mathrm{1}−{e}^{{ix}} \right)\:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}\left({n}+\mathrm{1}\right)}\:\:={Im}\left({S}\left({e}^{{ix}} \right)\right)\:\:{let}\:{find}\:{it} \\ $$$$\left({e}^{−{ix}} \:−\mathrm{1}\right)\:=\:{cosx}\:−{i}\:{sinx}\:−\mathrm{1} \\ $$$$ \\ $$$$=\:−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:−\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\:−\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:−{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=−\mathrm{2}{i}\:{e}^{−{i}\frac{{x}}{\mathrm{2}}} \\ $$$${ln}\left(\mathrm{1}−{e}^{{ix}} \right)\:=\:{ln}\:\left(\mathrm{1}−{cosx}\:−{isinx}\right) \\ $$$$=\:{ln}\left(\:\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:−\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$={ln}\left(−\mathrm{2}{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\left({cos}\left(\frac{{x}}{\mathrm{2}}\right)\:+{isin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\right. \\ $$$$=\:{ln}\left(−{i}\right)\:+{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{ln}\left({e}^{{i}\frac{{x}}{\mathrm{2}}} \right)\right. \\ $$$$=−{i}\:\frac{\pi}{\mathrm{2}}\:+{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\frac{{ix}}{\mathrm{2}} \\ $$$$={ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+{i}\frac{{x}−\pi}{\mathrm{2}} \\ $$$${Im}\left(\:{S}\left({e}^{{ix}} \right)\right)\:=\:−\mathrm{2}{i}\:{e}^{−{i}\frac{{x}}{\mathrm{2}}} \left(\:\:{ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\frac{{x}−\pi}{\mathrm{2}}\right)\right. \\ $$$$=−\mathrm{2}{i}\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\:−{isin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\left({ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\frac{{x}−\pi}{\mathrm{2}}\right)\right. \\ $$$$\left.=\:−\mathrm{2}{i}\:\right)\:\left(\:{cos}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\frac{{x}−\pi}{\mathrm{2}}\:{cos}\left(\frac{{x}}{\mathrm{2}}\right)\right.\right. \\ $$$$−{i}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+\frac{{x}−\pi}{\mathrm{2}}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$=\:−\mathrm{2}{i}\:{cos}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+\left({x}−\pi\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\right. \\ $$$$−\mathrm{2}\:{sin}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{i}\left(\pi−{x}\right)\:{sin}\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow\right. \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{n}\left({n}+\mathrm{1}\right)}\:\:=\left(\pi−{x}\right){sin}\left(\frac{{x}}{\mathrm{2}}\right)\:−{sin}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right) \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}\left({n}+\mathrm{1}\right)}\:=\left({x}−\pi\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)\:−\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\:+\mathrm{1}. \\ $$$$ \\ $$

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