Let $a$ be an odd integer $≥3$. It appears that: $$\lim_{n\rightarrow\infty}\frac{1}{2^{n}}\sum_{m=0}^{\left\lfloor n\frac{\ln2}{\ln a}\right\rfloor }\binom{n}{m}=\begin{cases} 1 & \textrm{if }a=3\\ 0 & \textrm{if }a\geq5 \end{cases}$$

Any ideas as to how to prove this? I tried using Stirling's formula, but everything became horrible.