# 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.

Mar 22, 2017

The series:

${\sum}_{n = 2}^{\infty} \ln \frac{n}{\sqrt{n}}$

is divergent.

#### Explanation:

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

$\ln n > 1$

and:

$\sqrt{n} < n$

so:

$\ln \frac{n}{\sqrt{n}} > \frac{1}{n} > 0$

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