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Question Number 41349 by math khazana by abdo last updated on 06/Aug/18
calculate Σ_(n=1) ^∞ (((−1)^n )/(3n−1))
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}−\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 06/Aug/18
let S = Σ_(n=1) ^∞ (((−1)^n )/(3n−1)) changement of indice n=j+1 give S =Σ_(j=0) ^∞ (((−1)^(j+1) )/(3j+2)) =−Σ_(j=0) ^∞ (((−1)^j )/(3j+2)) ⇒ S=−Σ_(n=0) ^∞ (((−1)^n )/(3n+2)) let w(x) =Σ_(n=0) ^∞ (((−1)^n )/(3n+2)) x^(3n+2) with ∣x∣<1 wehave S=w(1) and w^′ (x)=Σ_(n=0) ^∞ (−1)^n x^(3n+1) =x Σ_(n=0) ^∞ (−x^3 )^n =(x/(1+x^3 )) ⇒ w(x) = ∫ ((xdx)/(1+x^3 )) +c let decompse F(x)=(x/(1+x^3 )) =(x/((x+1)(x^2 −x+1))) F(x) = (a/(x+1)) +((bx+c)/(x^2 −x +1)) a =lim_(x→−1) (x+1)F(x)= ((−1)/3) lim_(x→+∞) xF(x) =0 =a+b ⇒b=(1/3) ⇒ F(x)=−(1/(3(x+1))) +(1/3) ((x+3c)/(x^2 −x+1)) F(0) = 0 =−(1/3) +c ⇒c=(1/3) ⇒F(x)=((−1)/(3(x+1))) +(1/3) ((x+1)/(x^2 −x+1)) ⇒ ∫ F(x)dx =−(1/3)ln∣x+1∣ +(1/6) ∫ ((2x−1+3)/(x^2 −x+1))dx =−(1/3)ln∣x+1∣ +(1/6)ln(x^2 −x+1) +(1/2) ∫ (dx/((x−(1/2))^2 +(3/4))) but ∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) (4/3) ∫ (1/(1+t^2 )) ((√3)/2) dt =(2/( (√3))) arctan(((2x−1)/( (√3))))⇒w(x)=−(1/3)ln∣x+1∣ +(1/6)ln(x^2 −x+1) +(2/( (√3)))arctan(((2x−1)/( (√3))))+c w(0)=0 =−(2/( (√3))) arctan((1/( (√3))))+c ⇒c=(2/( (√3)))arctan((1/( (√3))))=(2/( (√3)))(π/6) =(π/(3(√3))) ⇒ w(x) =−(1/3)ln∣x+1∣ +(1/6)ln(x^2 −x+1) +(2/( (√3))) arctan(((2x−1))/( (√3)))) +(π/(3(√3))) S=w(1)=−(1/3)ln(2) +(2/( (√3))) arctan((1/( (√3)))) +(π/(3(√3))) =−((ln(2))/3) +(2/( (√3))) (π/6) +(π/(3(√3))) S =((2π)/(3(√3))) −((ln(2))/3) .
$${let}\:{S}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}−\mathrm{1}}\:{changement}\:{of}\:{indice}\:{n}={j}+\mathrm{1}\:{give} \\ $$$${S}\:=\sum_{{j}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{j}+\mathrm{1}} }{\mathrm{3}{j}+\mathrm{2}}\:=−\sum_{{j}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{j}} }{\mathrm{3}{j}+\mathrm{2}}\:\Rightarrow\:{S}=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{2}}\:{let} \\ $$$${w}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{2}}\:{x}^{\mathrm{3}{n}+\mathrm{2}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\:{wehave}\:{S}={w}\left(\mathrm{1}\right)\:{and} \\ $$$${w}^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{3}{n}+\mathrm{1}} \:={x}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{3}} \right)^{{n}} \:=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\Rightarrow \\ $$$${w}\left({x}\right)\:=\:\int\:\:\frac{{xdx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\:+{c}\:\:{let}\:{decompse}\:{F}\left({x}\right)=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{3}} }\:=\frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)\:=\:\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\:\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:{F}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\:\frac{{x}+\mathrm{3}{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{−\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{3}}\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{6}}\:\int\:\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid\:\:+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:{but} \\ $$$$\int\:\:\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:\int\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{dt} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\Rightarrow{w}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c} \\ $$$${w}\left(\mathrm{0}\right)=\mathrm{0}\:=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c}\:\Rightarrow{c}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\frac{\pi}{\mathrm{6}}\:=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$${w}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid\:+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\left.\mathrm{2}{x}−\mathrm{1}\right)}{\:\sqrt{\mathrm{3}}}\right)\:+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${S}={w}\left(\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\frac{\pi}{\mathrm{6}}\:\:+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${S}\:=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 07/Aug/18
S=−w(1) ⇒ S=((ln(2))/3) −((2π)/(3(√3))) .
$${S}=−{w}\left(\mathrm{1}\right)\:\Rightarrow\:{S}=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{3}}\:−\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

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