Question-201888

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)

$${S}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\psi\left({n}+\mathrm{2}\right)}{\left({n}+\mathrm{2}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi\left({n}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} −\gamma}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\gamma\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right)\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\ast\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=−\zeta\left(\mathrm{3}\right)+\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\zeta\left(\mathrm{3}\right)+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} }{{n}^{\mathrm{2}} }=−\zeta\left(\mathrm{3}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$=−\zeta\left(\mathrm{3}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}=−\zeta\left(\mathrm{3}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$=−\zeta\left(\mathrm{3}\right)+\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right) \\ $$$$\Rightarrow{S}=\zeta\left(\mathrm{3}\right)−\gamma\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}\right) \\ $$

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