Question

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.

Answer 1

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.