Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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    .

$${let}\:{put}\:\:{s}\left({x}\right)=\:\:\sum_{{n}=\mathrm{0}} ^{{n}=\propto} \:\frac{{x}^{\mathrm{3}{n}+\mathrm{1}} }{\mathrm{3}{n}+\mathrm{1}}\:{with}\:\:/{x}/<\mathrm{1} \\ $$$${s}\left(^{\mathrm{3}} \sqrt{{x}\:}\right)\:\:=\:\underset{{n}=\mathrm{0}} {\sum}^{\propto} \:\:\frac{\left(^{\mathrm{3}} \sqrt{{x}}\right)^{\mathrm{3}{n}+\mathrm{1}} }{\mathrm{3}{n}+\mathrm{1}}\:\:=\:\:^{\mathrm{3}} \sqrt{{x}}\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:{and}\:{for}\:{x}\:\neq\mathrm{0} \\ $$$$\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:\:=\:\:^{\mathrm{3}} \left(\sqrt{{x}}\right)^{−\mathrm{1}} \:{s}\left(^{\mathrm{3}} \sqrt{\left.{x}\right)}\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:=−{s}\left(−\mathrm{1}\right)\right. \\ $$$${s}^{,} \left({x}\right)=\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:{x}^{\mathrm{3}{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{3}} }\Rightarrow\:?\:{s}\left({x}\right)=\:\lambda\:−\:\int\:\:\frac{{dx}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$$${after}\:{that}\:{we}\:{decompose}\:{F}\left({x}\right)=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{1}}\:=\:\:\:\frac{{a}}{{x}−\mathrm{1}}\:+\:\frac{{bx}+{c}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${we}\:{find}\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\left({x}−\mathrm{1}\right)}\:\:−\:\frac{{x}+\mathrm{2}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}\:\:{and} \\ $$$$\int{F}\left({x}\right){dx}=\:\frac{\mathrm{1}}{\mathrm{3}}{ln}/{x}−\mathrm{1}/\:−\:\frac{\mathrm{1}}{\mathrm{6}}{ln}/\:{x}^{\mathrm{2}} +{x}+\mathrm{1}/\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${s}\left({x}\right)=\lambda−\:\int/{F}\left({x}\right){dx}\Rightarrow{s}\left(\mathrm{0}\right)=\mathrm{0}=\:\lambda+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\ $$$$\lambda=−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:\:{and}\:{s}\left({x}\right)=\:−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:−\frac{\mathrm{1}}{\mathrm{3}}{ln}/{x}−\mathrm{1}/\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:{ln}/{x}^{\mathrm{2}} +{x}+\mathrm{1}/+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${finally}\:\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}=\:−{s}\left(−\mathrm{1}\right)=\:\:\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:{ln}\mathrm{2}\:\:\:\:. \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com