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n-0-1-n-3n-1-ln-2-1-3-

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Question Number 212164 by mnjuly1970 last updated on 04/Oct/24
Σ_(n=0) ^∞ (( (−1)^n )/(3n+1)) − ln((2)^(1/3) ) = ?
∑∞n=0(−1)n3n+1−ln(23)=?
Answered by Berbere last updated on 04/Oct/24
=Σ_(n≥0) ∫_0 ^1 (−x^3 )^n dx−((ln(2))/3)=∫_0 ^1 (1/(1+x^3 ))dx−((ln(2))/3) =∫_0 ^1 (1/((x+1)(x^2 −x+1)))dx−((ln(2))/3) =∫_0 ^1 (((x+1)(x−2)−(x^2 −x+1))/(−3(x+1)(x^2 −x+1)))dx−((ln(2))/3) =∫_0 ^1 ((x−2)/(−3(x^2 −x+1)))+(1/(3(x+1)))dx−((ln(2))/3) =(1/3)∫_0 ^1 ((2−x)/(x^2 −x+1)))dx=−(1/6)∫_0 ^1 ((2x−1)/(x^2 −x+1))dx+(1/2)∫_0 ^1 (dx/(x^2 −x+1)) =(1/2)∫_0 ^1 (dx/((x−(1/2))^2 +(3/4)))=(1/( (√3)))tan^(−1) (((2x)/( (√3)))−(1/( (√3))))]_0 ^1 =(2/( (√3))).(π/6)=(π/(3(√3)))
=∑n⩾0∫01(−x3)ndx−ln(2)3=∫0111+x3dx−ln(2)3=∫011(x+1)(x2−x+1)dx−ln(2)3=∫01(x+1)(x−2)−(x2−x+1)−3(x+1)(x2−x+1)dx−ln(2)3=∫01x−2−3(x2−x+1)+13(x+1)dx−ln(2)3=13∫012−xx2−x+1)dx=−16∫012x−1x2−x+1dx+12∫01dxx2−x+1=12∫01dx(x−12)2+34=13tan−1(2x3−13)]01=23.π6=π33
Commented by mnjuly1970 last updated on 04/Oct/24
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