sequences and series - Given a sufficiently large integer $N$, is it true that there are more primes than perfect squares in $[1,N]$?

We know that the sum of the reciprocals of all prime numbers diverges (that is, $\displaystyle\sum_{p\text { is prime}}p^{-1}=\infty$) and the sum of the reciprocals of all perfect squares converges (that is, $\displaystyle\sum_{k=1}^{\infty}k^{-2}<\infty$). Does this result imply that the set of prime numbers is more "dense" than the set of perfect squares; that is, given a sufficiently large integer $N$, there are always more primes than perfect squares in the closed interval $[1,N]$? If so, how about the answer to the following generalised question:

For a fixed positive number $\varepsilon$, does there exist a positive integer $N_{\varepsilon}$ such that if $N>N_{\varepsilon}$, there are always more primes than numbers of the form $k^{1+\varepsilon}$ (where $k$ is a positive number) in the closed interval $[1,N]$?

My guess is that the answer appears to be yes, since $\displaystyle\sum_{p\text { is prime}}p^{-1}=\infty$ whilst $\displaystyle\sum_{k=1}^{\infty}k^{-1-\varepsilon}<\infty$.