Question Number 144132 by islamo last updated on 21/Jun/21
Answered by ArielVyny last updated on 22/Jun/21
$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} }\backsim\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{{n}} }\:\left({CV}\right) \\ $$
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](https://www.tinkutara.com/question/Q144188.png)
$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} } \\ $$$${ln}\left({n}\right)^{{ln}\left({n}\right)} \geqslant{n}^{\mathrm{2}} \\ $$$$\Rightarrow{ln}\left({n}\right){ln}\left({lnn}\right)\geqslant\mathrm{2}{ln}\left({n}\right) \\ $$$$\Rightarrow{n}\geqslant{e}^{{e}^{\mathrm{2}} } \\ $$$$\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} }=\underset{{n}=\mathrm{2}} {\overset{\left[{e}^{{e}^{\mathrm{2}} } \right]} {\sum}}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} }+\underset{\left[{e}^{{e}^{\mathrm{2}} } \right]+\mathrm{1}} {\sum}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} } \\ $$$$={A}+{B},{B}\leqslant\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{0}\leqslant{S}<\zeta\left(\mathrm{2}\right)+\underset{\mathrm{2}\leqslant{n}\leqslant\left[{e}^{{e}^{\mathrm{2}} } \right]} {\sum}\frac{\mathrm{1}}{{ln}\left({n}\right)^{{ln}\left({n}\right)} } \\ $$$${Cv} \\ $$$$ \\ $$