How to determine the convergence of #Sigma_(n=2)^∞ lnn/sqrtn#?

Question

#Sigma_(n=2)^∞ lnn/sqrtn#
I need to prove this using a convergence/divergence test but I'm not sure how.

3314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-22T13:12:38

The series:

#sum_(n=2)^oo lnn/sqrtn#

is divergent.

We can use the direct comparison test: for #n>2# we have:

#ln n > 1#

and:

#sqrt n < n#

so:

#lnn/sqrtn > 1/n>0#

and as we know that the harmonic series is divergent, our series is also divergent.