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.
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.