How do you test the series #Sigma lnn/n# from n is #[1,oo)# for convergence?

1 Answer
Mar 9, 2017

The series:

#sum_(n=1)^oo lnn/n#

is divergent.

Explanation:

The function #lnx# is strictly increasing and as #ln e = 1# we have that #ln n > 1 # for #n > 3#.

Therefore:

#lnn/n > 1/n# for #n > 3#

and since #sum_(n=1)^oo 1/n# is a divergent series then also

#sum_(n=1)^oo lnn/n#

is divergent by direct comparison.