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Question Number 144132 by islamo last updated on 21/Jun/21

Answered by ArielVyny last updated on 22/Jun/21

Σ_(n≥0) (1/(ln(n)^(ln(n)) ))∽Σ_(n≥0) (1/n^n ) (CV)

∑n⩾01ln(n)ln(n)∽∑n⩾01nn(CV)

Answered by mindispower last updated on 22/Jun/21

Σ_(n≥2) (1/(ln(n)^(ln(n)) )) ln(n)^(ln(n)) ≥n^2 ⇒ln(n)ln(lnn)≥2ln(n) ⇒n≥e^e^2 Σ_(n≥2) (1/(ln(n)^(ln(n)) ))=Σ_(n=2) ^([e^e^2 ]) (1/(ln(n)^(ln(n)) ))+Σ_([e^e^2 ]+1) (1/(ln(n)^(ln(n)) )) =A+B,B≤Σ_(n=1) ^∞ (1/n^2 ) ⇒0≤S<ζ(2)+Σ_(2≤n≤[e^e^2 ]) (1/(ln(n)^(ln(n)) )) Cv

∑n⩾21ln(n)ln(n)ln(n)ln(n)⩾n2⇒ln(n)ln(lnn)⩾2ln(n)⇒n⩾ee2∑n⩾21ln(n)ln(n)=∑[ee2]n=21ln(n)ln(n)+∑[ee2]+11ln(n)ln(n)=A+B,B⩽∑∞n=11n2⇒0⩽S<ζ(2)+∑2⩽n⩽[ee2]1ln(n)ln(n)Cv

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