Perturbation Theory

a particle moves in one dimension in the potential

$V(x)=\infty \forall |x|>a, V(x)=V_0 \cos{\frac{\pi x}{2a}} \forall |x| \leq a$

now the unperturbed state that i use is just a standard infinite square well.

anyway the solution says that perturbation theory is only valid provided that the energy scale of the "bump" (set by $V_0$) is less than the difference in energy between square well states.
Q1: is this just a fact: perturbation theory is only applicable providing the perturbation is less than the energy difference of the states of teh unperturbed system?

it then explains the above mathematically by saying:

$V_0 << \frac{{\hbar}^2 \pi^2}{8ma^2}(2n-1)$
i can't for the life of me see where the RHS of that comes from. arent the energy levels dependent on $n^2$ and not $n$?

$$n^2- (n-1)^2 = n^2 - (n^2 - 2n+1) = 2n-1$$