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Question Number 183967 by SEKRET last updated on 01/Jan/23
Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1)) = ?
∑∞n=1(−1)n+13n−1=?
Answered by Ar Brandon last updated on 01/Jan/23
S=Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=Σ_(n=1) ^∞ (((−1)^(3n−1) )/(3n−1)) S(t)=Σ_(n=1) ^∞ (t^(3n−1) /(3n−1)) ⇒S ′(t)=Σ_(n=1) ^∞ t^(3n−2) =(t/(1−t^3 )) ⇒S(t)=∫(t/(1−t^3 ))dt=−∫(t/((t−1)(t^2 +t+1)))dt ⇒S(t)=−∫((1/(3(t−1)))−((t−1)/(3(t^2 +t+1))))dt ⇒S(t)=−(1/3)ln∣t−1∣+(1/6)∫((2t+1)/(t^2 +t+1))dt−(1/2)∫(1/(t^2 +t+1))dt =−(1/3)ln∣t−1∣+(1/6)ln(t^2 +t+1)−(1/( (√3)))arctan(((2t+1)/( (√3))))+C S(0)=0=−(1/( (√3)))arctan((1/( (√3))))+C ⇒C=(π/(6(√3))) ⇒S(t)=−(1/( 3))ln∣t−1∣+(1/6)ln(t^2 +t+1)−(1/( (√3)))arctan(((2t+1)/( (√3))))+(π/(6(√3))) ⇒Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=S(−1)=−((ln2)/3)−(1/( (√3)))arctan(−(1/( (√3))))+(π/(6(√3))) ⇒ determinant (((Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−1))=−((ln2)/3)+(π/(3(√3))))))
S=∑∞n=1(−1)n+13n−1=∑∞n=1(−1)3n−13n−1S(t)=∑∞n=1t3n−13n−1⇒S′(t)=∑∞n=1t3n−2=t1−t3⇒S(t)=∫t1−t3dt=−∫t(t−1)(t2+t+1)dt⇒S(t)=−∫(13(t−1)−t−13(t2+t+1))dt⇒S(t)=−13ln∣t−1∣+16∫2t+1t2+t+1dt−12∫1t2+t+1dt=−13ln∣t−1∣+16ln(t2+t+1)−13arctan(2t+13)+CS(0)=0=−13arctan(13)+C⇒C=π63⇒S(t)=−13ln∣t−1∣+16ln(t2+t+1)−13arctan(2t+13)+π63⇒∑∞n=1(−1)n+13n−1=S(−1)=−ln23−13arctan(−13)+π63⇒∑∞n=1(−1)n+13n−1=−ln23+π33
Commented by Ar Brandon last updated on 01/Jan/23
★★Happy New Year friends ★★
★★HappyNewYearfriends★★
Commented by SEKRET last updated on 01/Jan/23
thanks sir
thankssir
Commented by SEKRET last updated on 01/Jan/23
thanks
thanks
Answered by witcher3 last updated on 01/Jan/23
=−Σ_(n≥1) ∫_0 ^1 (−1)^(n+1) x^(3n−2) dx =−∫_0 ^1 (1/x^2 )Σ_(n≥1) (−x^3 )^n dx =−∫_0 ^1 ((−x)/(1+x^3 ))dx=−∫_0 ^1 (x/((x+1)(x^2 −x+1)))dx =∫_0 ^1 (((x+1)^2 −(x^2 −x+1))/((x+1)(x^2 −x+1)))dx ={∫_0 ^1 ((x+1)/(x^2 −x+1))−ln(2)} =∫_0 ^1 ((2x−1)/(2(x^2 −x+1)))dx+(3/2)∫_0 ^1 (dx/(x^2 −x+1))−ln(2) =−ln(2)+(3/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4))) =−ln(2)+2∫_0 ^1 (dx/(1+(((2x)/( (√3)))−(1/( (√3))))^2 )) =−ln(2)+(√(3.))2.tan^(−1) ((1/( (√3)))) =−ln(2)+(π/( (√3)))
=−∑n⩾1∫01(−1)n+1x3n−2dx=−∫011x2∑n⩾1(−x3)ndx=−∫01−x1+x3dx=−∫01x(x+1)(x2−x+1)dx=∫01(x+1)2−(x2−x+1)(x+1)(x2−x+1)dx={∫01x+1x2−x+1−ln(2)}=∫012x−12(x2−x+1)dx+32∫01dxx2−x+1−ln(2)=−ln(2)+32∫01dx(x−12)2+34=−ln(2)+2∫01dx1+(2x3−13)2=−ln(2)+3.2.tan−1(13)=−ln(2)+π3
Commented by Ar Brandon last updated on 01/Jan/23
(x+1)^2 −(x^2 −x+1)=3x
(x+1)2−(x2−x+1)=3x
Commented by SEKRET last updated on 01/Jan/23
thanks sir
thankssir

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