In this paper, the authors say that for any $\gamma \in [1/2,1)$, there is a positive constant $A=A(\gamma)$ such that for any $n$,
$$ \sum_{n\gamma\leq k \leq n} \binom{n}{k} \geq A n^{-1/2}2^{n \cdot h(\gamma)}, $$ where $h(x)=-x\log_{2} x - (1-x)\log_{2}(1-x)$.
They also say that that fact follows immediately from Stirling's formula. My question is why does this follow from Stirling's formula?
I'd prefer a push in the right direction to a full-fledged solution.
This is related to this question.