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

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

1 Answer
Mar 22, 2017

The series:

sum_(n=2)^oo lnn/sqrtn

is divergent.

Explanation:

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.