Let $\displaystyle f(n,k) = \sum_{m \le k} \binom{n}{m}$. We need to show $\displaystyle f(n,k) \sim \frac{2^{nH(k/n)}}{\sqrt{2\pi k(n-k)/n}}(1 + o(1))$, where $k = o(n)$ and $k \to \infty$
Actually I've thought about three ideas :
1) Consider $\displaystyle \xi _{i}$ be Bernoulli random variables, and use CLT for estimating this sum.
2) Use Stirling approximation.
3) Use Abel summation.
The former idea give the worst asymptotic. The second one isn't usable , because $m$ is just a constant for finitely many times.
The last idea gives me :
(1) = $\displaystyle \sum_{m \le k} \binom{n}{m} = \sum \frac{n!}{m!} \cdot \frac{1}{(n-m)!}$, let $\displaystyle g(x) = \frac{1}{\Gamma(n-x)}$ and $a_m = n! / m!$. Hence we have $\displaystyle A(x) = \sum_{m \le x} \frac{n!}{m!} \sim n! e$, so we would have something like (1) $\displaystyle \sim$ $\displaystyle \frac{en!}{\Gamma(n-k)} + \int_{1}^{k} \frac{en! \Gamma'(n-x)}{\Gamma^2(n-x)} dx$. Major part of integral gives us $\displaystyle \frac{en!}{\Gamma(n-k)}$, so we obtain $\displaystyle \frac{2e n!}{\Gamma(n-k)} = \frac{2en!}{(n-k)!}$ even Stirling approximation doesn't give us the mentioned result. Any hint?
Attempt (using CLT). Let $S_n = \sum X_i$, where $X_i$ i.r.v with Bernoulli distribution and $p = 1/2$. Hence we have $2^n \mathbb{P}( S_n \le k) = \sum ^{k} \binom{n}{m}$. So we should estimate this probability.
Using CLT we have : $ \mathbb{P}\left( \frac{S_n - \mathbb{E}(S_n) }{\sqrt{\mathbb{Var}(S_n)}} \le \frac{k - \mathbb{E}(S_n)}{ \sqrt{\mathbb{Var}(S_n)}}\right)$. Thus for $n \to \infty$ we have that probability estimates as $F_{Z} \left( \frac{k - n/2}{ \sqrt{n/4}}\right)$ which doesnt give us a given estimation. Maybe this approach can be obtained and upgraded?