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

1 Answer
Mar 9, 2017

The series:

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

is convergent.

Explanation:

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

L= lim_(n->oo) root(n)(a_n) = lim_(n->oo) root(n)((ln(n)^(-n))) = lim_(n->oo) 1/lnn = 0

As L < 1 the series is convergent.