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

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Answer 1 · 2017-03-09T00:27:28

The series:

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

is divergent.

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.