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Question Number 201888 by MrGHK last updated on 15/Dec/23

Answered by namphamduc last updated on 15/Dec/23

S=Σ_(n=0) ^∞ ((ψ(n+2))/((n+2)^2 ))=Σ_(n=1) ^∞ ((ψ(n+1))/((n+1)^2 ))=Σ_(n=1) ^∞ ((H_n −γ)/((n+1)^2 )) =−γ((π^2 /6)−1) +Σ_(n=1) ^∞ (H_n /((n+1)^2 )) ∗Σ_(n=1) ^∞ (H_n /((n+1)^2 ))=Σ_(n=1) ^∞ ((H_(n+1) −(1/(n+1)))/((n+1)^2 ))=−ζ(3)+1+Σ_(n=1) ^∞ (H_(n+1) /((n+1)^2 )) =−ζ(3)+Σ_(n=1) ^∞ (H_n /n^2 )=−ζ(3)−Σ_(n=1) ^∞ (1/n)∫_0 ^1 x^(n−1) ln(1−x)dx =−ζ(3)+∫_0 ^1 ((ln^2 (1−x))/x)dx=−ζ(3)+∫_0 ^1 ((ln^2 (x))/(1−x))dx =−ζ(3)+2Σ_(n=0) ^∞ (1/((n+1)^3 ))=ζ(3) ⇒S=ζ(3)−γ((π^2 /6)−1)

S=∑∞n=0ψ(n+2)(n+2)2=∑∞n=1ψ(n+1)(n+1)2=∑∞n=1Hn−γ(n+1)2=−γ(π26−1)+∑∞n=1Hn(n+1)2∗∑∞n=1Hn(n+1)2=∑∞n=1Hn+1−1n+1(n+1)2=−ζ(3)+1+∑∞n=1Hn+1(n+1)2=−ζ(3)+∑∞n=1Hnn2=−ζ(3)−∑∞n=11n∫01xn−1ln(1−x)dx=−ζ(3)+∫01ln2(1−x)xdx=−ζ(3)+∫01ln2(x)1−xdx=−ζ(3)+2∑∞n=01(n+1)3=ζ(3)⇒S=ζ(3)−γ(π26−1)

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