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.



Any ideas for the following limit of partial sums of binomial coefficients?
Fix $0<\alpha<\frac12$ and consider the sum $\sum\_{0\le m\le\alpha n} \binom nm$. (For simplicity let's assume $\alpha n$ is an integer.) The ratio of the $m$th term to the $(m+1)$st term in this sum is at most $\alpha/(1-\alpha)$; this means that the sum is bounded by
$$
\binom n{\alpha n} \sum\_{k=0}^\infty \bigg(\frac\alpha{1-\alpha}\bigg)^k = \binom n{\alpha n} \frac{\alpha-1}{2\alpha-1},
$$
which is $o(2^n)$ by Stirling's formula or probably a simpler argument (an upper bound for the central binomial coefficient will suffice). A similar argument works for the upper tail $\sum\_{\beta n\le m\le n} \binom nm$ when $\frac12<\beta<1$.


