Question Number 102544 by 175mohamed last updated on 09/Jul/20

Answered by Rio Michael last updated on 10/Jul/20

for p > 1, Σ_(n=1) ^∞ (1/n^p ) is convergent ⇒ Σ_(n=1) ^∞ (1/(n^2 )^(1/3) ) diverges

Answered by mathmax by abdo last updated on 10/Jul/20

S =Σ_(n=1) ^∞  (1/n^(2/3) )   the function ϕ(t) =(1/t^(2/3) )  are >0 decreazing so S have same  nature with ∫_1 ^∞  (dt/t^(2/3) )  this integral is divergent ⇒S is divergent  any way Σ (1/n^α ) converge for n>1 and diverges for α≤1

Commented by175mohamed last updated on 13/Jul/20