Originally Posted by
mr fantastic Your use of the comparison test is fine:
$\displaystyle \frac{\sqrt{k} 4^k}{2 + k 5^k} < \frac{k 4^k}{k 5^k} = \frac{4^k}{5^k} = \left( \frac{4}{5} \right)^k$.
Without numerical number crunching, I'm not sure how to establish how large n should be.