How do you use the integral test to determine if #sum_(n=2)^oo 1/(nsqrtlnn)# from #[2,oo)# is convergent or divergent?

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Answer 1 · 2016-11-21T16:01:47

#sum_(k=2)^n1/(k log_e(k))# is divergent

Calling #f(x)=1/(x log_e(x))# we have that

#int_2^n f(x)dx le sum_(k=2)^n1/(k log_e(k))# and

#int_2^n f(x)dx=log_e((log_e n)/(log_e 2))# but

#lim_(n->oo)log_e(n)=oo# so

#lim_(n->oo)sum_(k=2)^n1/(k log_e(k))=oo#