Question

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

Answer 1

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

Explanation:

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`