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Question Number 26565 by abdo imad last updated on 26/Dec/17
find the value of Σ_(n=0) ^∝ (((−1)^n )/(3n+1))
$${find}\:{the}\:{value}\:{of}\:\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$
Commented by abdo imad last updated on 29/Dec/17
let put s(x)= Σ_(n=0) ^(n=∝) (x^(3n+1) /(3n+1)) with /x/<1 s(^3 (√(x ))) = Σ^∝ _(n=0) (((^3 (√x))^(3n+1) )/(3n+1)) = ^3 (√x) Σ_(n=0) ^∝ (x^n /(3n+1)) and for x ≠0 Σ_(n=0) ^∝ (x^n /(3n+1)) = ^3 ((√x))^(−1) s(^3 (√(x))) and Σ_(n=0) ^∝ (((−1)^n )/(3n+1)) =−s(−1) s^, (x)= Σ_(n=0) ^∝ x^(3n) = (1/(1−x^3 ))⇒ ? s(x)= λ − ∫ (dx/(x^3 −1)) after that we decompose F(x)= (1/(x^3 −1)) = (a/(x−1)) + ((bx+c)/(x^2 +x+1)) we find F(x) = (1/(3(x−1))) − ((x+2)/(3(x^2 +x+1))) and ∫F(x)dx= (1/3)ln/x−1/ − (1/6)ln/ x^2 +x+1/ +((√3)/3) arctan((2/( (√3)))(x+(1/2))) s(x)=λ− ∫/F(x)dx⇒s(0)=0= λ+ ((√3)/3) arctan((1/( (√3)))) λ=−((π(√3))/(18)) and s(x)= −((π(√3))/(18)) −(1/3)ln/x−1/ + (1/6) ln/x^2 +x+1/+ ((√3)/3) arctan((2/( (√3)))(x+(1/2))) finally Σ_(n=0) ^∝ (((−1)^n )/(3n+1))= −s(−1)= ((π(√3))/9) + (1/3) ln2 .
$${let}\:{put}\:\:{s}\left({x}\right)=\:\:\sum_{{n}=\mathrm{0}} ^{{n}=\propto} \:\frac{{x}^{\mathrm{3}{n}+\mathrm{1}} }{\mathrm{3}{n}+\mathrm{1}}\:{with}\:\:/{x}/<\mathrm{1} \\ $$$${s}\left(^{\mathrm{3}} \sqrt{{x}\:}\right)\:\:=\:\underset{{n}=\mathrm{0}} {\sum}^{\propto} \:\:\frac{\left(^{\mathrm{3}} \sqrt{{x}}\right)^{\mathrm{3}{n}+\mathrm{1}} }{\mathrm{3}{n}+\mathrm{1}}\:\:=\:\:^{\mathrm{3}} \sqrt{{x}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:{and}\:{for}\:{x}\:\neq\mathrm{0} \\ $$$$\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:\:=\:\:^{\mathrm{3}} \left(\sqrt{{x}}\right)^{−\mathrm{1}} \:{s}\left(^{\mathrm{3}} \sqrt{\left.{x}\right)}\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:=−{s}\left(−\mathrm{1}\right)\right. \\ $$$${s}^{,} \left({x}\right)=\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:{x}^{\mathrm{3}{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{3}} }\Rightarrow\:?\:{s}\left({x}\right)=\:\lambda\:−\:\int\:\:\frac{{dx}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$$${after}\:{that}\:{we}\:{decompose}\:{F}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}\:=\:\:\:\frac{{a}}{{x}−\mathrm{1}}\:+\:\frac{{bx}+{c}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${we}\:{find}\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\left({x}−\mathrm{1}\right)}\:\:−\:\frac{{x}+\mathrm{2}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}\:\:{and} \\ $$$$\int{F}\left({x}\right){dx}=\:\frac{\mathrm{1}}{\mathrm{3}}{ln}/{x}−\mathrm{1}/\:−\:\frac{\mathrm{1}}{\mathrm{6}}{ln}/\:{x}^{\mathrm{2}} +{x}+\mathrm{1}/\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${s}\left({x}\right)=\lambda−\:\int/{F}\left({x}\right){dx}\Rightarrow{s}\left(\mathrm{0}\right)=\mathrm{0}=\:\lambda+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\lambda=−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:\:{and}\:{s}\left({x}\right)=\:−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}/{x}−\mathrm{1}/\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:{ln}/{x}^{\mathrm{2}} +{x}+\mathrm{1}/+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${finally}\:\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\:−{s}\left(−\mathrm{1}\right)=\:\:\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:{ln}\mathrm{2}\:\:\:\:. \\ $$$$ \\ $$

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