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Question-61907




Question Number 61907 by naka3546 last updated on 11/Jun/19
Commented by MJS last updated on 12/Jun/19
Σ_(n=1) ^∞ (1/n)=∞ ⇒ Σ_(n=1) ^∞ (1/( (n)^(1/k) ))=∞ for k∈N^★     calculating Ω_n  I get:  Ω_(125) ≈.849696  Ω_(250) ≈.846055  Ω_(500) ≈.843624  Ω_(1000) ≈.842011  Ω_(2000) ≈.840945
n=11n=n=11nk=forkNcalculatingΩnIget:Ω125.849696Ω250.846055Ω500.843624Ω1000.842011Ω2000.840945
Answered by tanmay last updated on 11/Jun/19
lim_(n→∞)  (N_r /D_r )  N_r =[((1/1))^(1/5) +((1/2))^(1/5) +((1/3))^(1/5) +...+((1/n))^(1/5) ]^(2/3)   D_r =[((1/1))^(1/3) +((1/2))^(1/3) +((1/3))^(1/3) +...+((1/n))^(1/3) ]^(4/5)   N_r =[((1/1))^(1/5) +((1/2))^(1/5) +((1/3))^(1/5) +...+((1/n))^(1/5) ]^((10)/(15))   D_r =[((1/1))^(1/3) +((1/2))^(1/3) +((1/3))^(1/3) +...+((1/n))^(1/3) ]^((12)/(15))   N_r =[{((1/1))^(1/5) +((1/2))^(1/5) +((1/3))^(1/5) +...+((1/n))^(1/5) }^5 ]^(2/(15))   =Σ_(n=1) ^n [{((1/n))^(1/5) }^5 ]^(2/(15))  and D_r =Σ_(n=1) ^∞ [{((1/n))^(1/3) }^6 ]^(2/(15))   N_r =[(1/1)+(1/2)+..+(1/n)+g(n)]^(2/(15))   D_r =[(1/1^2 )+(1/2^2 )+..+(1/n^2 )+h(n)]^(2/(15))   (1/n^2 )<(1/n)    D_r <N_r   wait...
limnNrDrNr=[(11)15+(12)15+(13)15++(1n)15]23Dr=[(11)13+(12)13+(13)13++(1n)13]45Nr=[(11)15+(12)15+(13)15++(1n)15]1015Dr=[(11)13+(12)13+(13)13++(1n)13]1215Nr=[{(11)15+(12)15+(13)15++(1n)15}5]215=nn=1[{(1n)15}5]215andDr=n=1[{(1n)13}6]215Nr=[11+12+..+1n+g(n)]215Dr=[112+122+..+1n2+h(n)]2151n2<1nDr<Nrwait

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