In page 3 of the paper On a new law of large numbers, by Erdos and Renyi one reads:
where $$h(x) = x \log_2 \frac{1}{x}+(1-x) \log_2 \frac{1}{1-x}.$$
Typo?
There might be a typo here. Take $\gamma =1/2$ by the symmetry of the binomial coefficients together with $\sum_k {n\choose k} = 2^n$ we get that $$\sum_{n/2\leq K\leq n} 2^{-n} {n\choose k} \geq 1/2 $$
Now since $h(1/2) = 1$. The upper bound inequality in (2.6) of the paper becomes:
$$1/2\leq \sum_{n/2\leq K\leq n} 2^{-n} {n\choose k} \leq B_1n^{-1/2} \to 0 $$
which is in contradiction.
Is there an error on (2.6) of the paper?
Question
If we required $\gamma$ to be strictly bigger than $1/2$ would such inequality hold?
How do we use Stirling formula in this case?
$${n \choose K} = \frac{n!}{(n - K)!K!} \sim \sqrt{\frac{n}{2\pi(n-K)K}}\bigg(\frac{n}{n-K}\bigg)^{n-K}\bigg(\frac{n}{K}\bigg)^{K}$$
But now, how do we sum this terms to get the desired inequalities (upper and lower bounds?)
The point is that, if $\gamma$ is a fixed contant greater than $1/2$, then $$\sum\limits_{\gamma n \le k \le n} \binom{n}{k} = \Theta\left(\binom{n}{\gamma n}\right).$$ This is because each next binomial coefficient is constant-time smaller than the previous one: $$\frac{\binom{n}{k + 1}}{\binom{n}{k}} = \frac{n - k}{k + 1} \le \frac{1 - \gamma}{\gamma} < 1,$$ for $k \ge \gamma n$. Thus we can only analize the first coefficient in the sum. This can be done directly by Stirling's formula: $$\binom{n}{\gamma n} \sim \sqrt{\frac{1}{2\pi\gamma (1 - \gamma) n}} \cdot 2^{h(\gamma)n},$$ as $n\to\infty$, and (2.6) follows (once again, for every fixed $\gamma > 1/2$).