In this paper, the authors say that for any $\gamma \in [1/2,1)$, there is a positive constant $B=B(\gamma)$ such that for any $n$,
$$ \sum_{n\gamma\leq k \leq n} \binom{n}{k} \leq B n^{-1/2}2^{n \cdot h(\gamma)}, $$ 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?
Here's what I've tried: Stirling's formula says that $n! = (1+R(n))\sqrt{2\pi n}n^{n}e^{-n}$ for some positive sequence $R(n)$ that tends to $0$ as $n\to \infty$.
By just replacing the three factorials in $\binom{n}{k}$ (i.e. the $n!$, the $k!$ and the $(n-k)!$) by expressions coming from Stirling's formula, I get
$$ \sum_{n\gamma \leq k \leq n} \binom{n}{k} = n^{-1/2}(2\pi)^{-1/2} \sum_{n\gamma \leq k \leq n} \frac{1+R(n)}{(1+R(k))(1+R(n-k))}\frac{1}{\sqrt{(k/n)(1-k/n)}}2^{nh(k/n)} $$
Using the fact that $R(n)$ is positive and convergent, the $\frac{1+R(n)}{(1+R(k))(1+R(n-k))}$ is bounded by some positive constant $R$. Basic properties of the function $h(x)$ also show that $h(k/n) \leq h(\gamma)$, so the $2^{nh(k/n)}$ is bounded by $2^{nh(\gamma)}$, as desired. But now what's left is to bound $$ \sum_{n\gamma \leq k \leq n}\frac{1}{\sqrt{(k/n)(1-k/n)}}, $$ which is impossible since when $n$ is large, the $k=n-1$ term will make that sum arbitrarily large. What am I missing? The problem seems to be related to the fact that Stirling's formula (as I stated it) isn't valid for $n=0$, but I can't see how to get around this.
Hint: For every $0\leqslant k\leqslant i\leqslant n$, $${n\choose i}={n\choose k}\cdot\prod_{j=k}^{i-1}\frac{n-j}{j+1}\leqslant{n\choose k}\cdot\left(\frac{n-k}{k+1}\right)^{i-k},$$ hence, if $2k\geqslant n$, summing the geometric series with reason $$r=\frac{n-k}{k+1}\lt1,$$ one gets $$\sum_{i=k}^n{n\choose i}\lt{n\choose k}\cdot\sum_{i\geqslant k}r^{i-k}={n\choose k}\cdot\frac{k+1}{2k+1-n}.$$ Now, for every $\gamma\gt\frac12$, use this for $$k=\lceil\gamma n\rceil,$$ with the guarantee that $2k\geqslant n$ since $\gamma\geqslant\frac12$, and apply Stirling formula to the binomial prefactor $${n\choose \lceil\gamma n\rceil}.$$