Question Number 31981 by abdo imad last updated on 17/Mar/18
$${find}\:{the}\:{nature}\:{of}\:\:\:\sum_{{n}\geqslant\mathrm{2}} \:\frac{\mathrm{1}}{{nln}\left({n}\right)}\:. \\ $$
Commented by abdo imad last updated on 22/Mar/18
$${the}\:{sequence}\:{u}_{{n}} =\:\frac{\mathrm{1}}{{nln}\left({n}\right)}\:{is}\:{decreasing}\:{with}\:{u}_{{n}} \geqslant{o}\:{so} \\ $$$$\Sigma\:{u}_{{n}} \:{and}\:\int_{\mathrm{2}} ^{+\infty} \:\frac{{dx}}{{xln}\left({x}\right)}\:{have}\:{thesame}\:{nature}\:{let}\:{prove} \\ $$$${tbe}\:{divervence}\:{of}\:{this}\:{integral}\:.{ch}.{ln}\left({x}\right)={t}\:{give} \\ $$$$\int_{\mathrm{2}} ^{+\infty} \:\:\:\frac{{dx}}{{xln}\left({x}\right)}\:=\:\int_{{ln}\left(\mathrm{2}\right)} ^{\infty} \:\:\frac{{e}^{{t}} {dt}}{{t}\:{e}^{{t}} }\:=\int_{{ln}\left(\mathrm{2}\right)} ^{+\infty} \:\frac{{dt}}{{t}}\:{diverges}\:{so} \\ $$$$\Sigma{u}_{{n}} \:\:{diverges}. \\ $$