How do you test for convergence of #sum_(n=2)^(oo) lnn^(-lnn)#?

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Answer 1 · 2017-05-26T20:12:17

#sum_(n=2)^oo ln(n^(-ln n)) = -oo#

Based on the properties of logarithms:

#ln(n^(-ln n)) = (-ln n) (ln n) = -(ln n)^2#

This means that:

#lim_(n->oo) a_n = lim_(n->oo) -(lnn)^2 = -oo#

The series does not satisfy Cauchy's necessary condition:

#lim_(n->oo) a_n = 0#

so it is not convergent.

As the terms are all negative it is divergent and:

#sum_(n=2)^oo ln(n^(-ln n)) = -oo#