Evaluate fraction of sumSo i have to evaluate this sum:1−2−2+4−2−5−2+7−2−8−2+10−2−11−2+⋯1+2−2−4−2−5−2+7−2+8−2−10−2−11−2+⋯it has the form : ∑0∞[(3n+1)−2−(3n+2)−2]∑0∞(−1)n[(3n+1)−2+(3n+2)−2]My current attempt : Trying to convert this into power seriesa(n)=(3n+1)−2−(3n+2)−2 b(n)=(3n+1)−2+(3n+2)−2Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ?Maybe there is a better direction. Can someone give me a hint ?

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Evaluate fraction of sumSo i have to evaluate this sum:1−2−2+4−2−5−2+7−2−8−2+10−2−11−2+⋯1+2−2−4−2−5−2+7−2+8−2−10−2−11−2+⋯it has the form : ∑0∞[(3n+1)−2−(3n+2)−2]∑0∞(−1)n[(3n+1)−2+(3n+2)−2]My current attempt : Trying to convert this into power seriesa(n)=(3n+1)−2−(3n+2)−2      b(n)=(3n+1)−2+(3n+2)−2Can a(n) and b(n) be the definite integral of certain polynomial function f(x) ?Maybe there is a better direction. Can someone give me a hint ?

cineworld93uowb · Accepted Answer

LetLi2(z)=∑n=1∞znn2,Cl2(θ)=∑n=1∞sin⁡nθn2=Im[Li2(eiθ)]be dilogarithm and Clausen function, respectively. Then it is easy to see that the expression in question reduces toCl2(2π3)Cl2(π3).By comparing the power series of both sides, we obtainLi2(z)+Li2(−z)=12Li2(z2).Now plugging z=eπi3 givesLi2(e2πi3)=2Li2(eπi3)+2Li2(−eπi3).Now taking imaginary part, we obtainCl2(2π3)=2Cl2(π3)−2Cl2(2π3),since −eπi3=e−2πi3.Therefore we haveCl2(2π3)Cl2(π3)=23.

Jambrichp2w2 · Accepted Answer

First, let  A=1−122+142−152+… and B=1+122−142−152+… Notice that the both A and B converges absolutely (by comparison test). Thus, limit law applies: B−A=2×122−2×142−2×182−2×1102+…=A2 and AB=23Lol i was so naive back then.

Evaluate fraction of sum So i have to evaluate

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