# How do you test for convergence of Sigma (ln(n))^-n from n=[2,oo)?

Mar 9, 2017

The series:

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

is convergent.

#### Explanation:

The series has positive terms so we can use the root test and we have:

$L = {\lim}_{n \to \infty} \sqrt[n]{{a}_{n}} = {\lim}_{n \to \infty} \sqrt[n]{\left(\ln {\left(n\right)}^{- n}\right)} = {\lim}_{n \to \infty} \frac{1}{\ln} n = 0$

As $L < 1$ the series is convergent.