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

May 26, 2017

${\sum}_{n = 2}^{\infty} \ln \left({n}^{- \ln n}\right) = - \infty$

#### Explanation:

Based on the properties of logarithms:

$\ln \left({n}^{- \ln n}\right) = \left(- \ln n\right) \left(\ln n\right) = - {\left(\ln n\right)}^{2}$

This means that:

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} - {\left(\ln n\right)}^{2} = - \infty$

The series does not satisfy Cauchy's necessary condition:

${\lim}_{n \to \infty} {a}_{n} = 0$

so it is not convergent.

As the terms are all negative it is divergent and:

${\sum}_{n = 2}^{\infty} \ln \left({n}^{- \ln n}\right) = - \infty$