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

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#Sigma_(n=2)^∞ lnn/sqrtn#

I need to prove this using a convergence/divergence test but I'm not sure how.

I need to prove this using a convergence/divergence test but I'm not sure how.

##### 1 Answer

The series:

is divergent.

#### Explanation:

We can use the direct comparison test: for

and:

so:

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