What is the asymptotic behavior of $\sum_{k\in B_n(a)}\binom{n}{k}$? Inspired by an answer to this or a comment to this question, I proved that if $a\in(0,1)$ and $A_n(a):=\{0,1,...,n\}\cap\left[\frac{1-a}{2}n,\frac{1+a}{2}n\right]$ resp. $B_n(a):=\{0,1,...,n\}\backslash A_n(a)$ we have
$$
\lim_{n\to\infty}\frac{1}{2^n}\sum_{k\in A_n(a)}\binom{n}{k}=1 \qquad\text{and}\qquad \lim_{n\to\infty}\frac{1}{2^n}\sum_{k\in B_n(a)}\binom{n}{k}=0.
$$
For this I showed that
$$
\sum_{k\in B_n(a)}\binom{n}{k}≤\left(\frac{2}{\sqrt{(1-a)^{1-a}(1+a)^{1+a}}}\right)^n
$$
which suffices since
$$
(1-a)^{1-a}(1+a)^{1+a}>1.
$$
However, to understand the behavior of $\sum_{k\in B_n(a)}\binom{n}{k}$ I calculated a few values, which let me to conjecture that
$$
\lim_{n\to\infty}{\sqrt{n}\left(\frac{\sqrt{(1-a)^{1-a}(1+a)^{1+a}}}{2}\right)^n}\sum_{k\in B_n(a)}\binom{n}{k}=\frac{1}{a}.
$$
Is this true? If yes, how to prove it? If no, how to precisely describe the behavior of $\sum_{k\in B_n(a)}\binom{n}{k}$?
Edit:
As Did pointed out, the value of ${\sqrt{n}\left(\frac{\sqrt{(1-a)^{1-a}(1+a)^{1+a}}}{2}\right)^n}\sum_{k\in B_n(a)}\binom{n}{k}$ jumps if $\frac{1\pm a}{2}n$ crosses an integer, which makes them unlikely to converge. However, the values seem always to be somewhat close to $\frac{1}{a}$, even for big $n$. Can this in some sort be explained or even quantified?
 A: 
Summary: The convergence you are asking about does not hold. The normalization you suggest is the correct one in the sense that  its limsup is positive and finite at least when $a$ is rational, even though its liminf is always $0$.

To prove this, first define $$S_n(a)=\sqrt{n}\,\alpha(a)^{-n}\sum_{k\in B_n(a)}{n\choose k}$$ where $$\alpha(a)=\frac12\sqrt{(1+a)^{1+a}(1-a)^{1-a}}$$
Next, consider $r_n(a)=\frac1n\lceil\frac{1+a}2n\rceil$ and $s_n(a)=1-r_n(a)$ and note that, when $n\to\infty$, $r_n(a)\to\frac{1+a}2$ and $s_n(a)\to\frac{1-a}2$, hence Stirling formula yields the exact equivalent $${n\choose r_n(a)n}\sim\frac{e^{-nt_n(a)}}{\sqrt{2\pi r_n(a)s_n(a)n}}$$ with $$t_n(a)=r_n(a)\log r_n(a)+s_n(a)\log s_n(a)$$
Note that $r_n(a)=\frac{1+a}2+\frac{e_n(a)}n$ with $0\leqslant e_n(a)<1$ hence $$r_n(a)\log r_n(a)=\left(\frac{1+a}2+\frac{e_n(a)}n\right)\left(\log\left(\frac{1+a}2\right)+\frac{2e_n(a)}{(1+a)n}+O\left(\frac1{n^2}\right)\right)$$
that is, $$r_n(a)\log r_n(a)=\frac{1+a}2\log\left(\frac{1+a}2\right)+e_n(a)\log\left(\frac{1+a}2\right)\frac1n+\frac{e_n(a)}n+O\left(\frac1{n^2}\right)$$
Likewise,  $s_n(a)=\frac{1-a}2-\frac{e_n(a)}n$ hence, replacing $a$ by $-a$ and $e_n(a)$ by $-e_n(a)$ in the expansion of $r_n(a)\log r_n(a)$, one gets $$s_n(a)\log s_n(a)=\frac{1-a}2\log\left(\frac{1-a}2\right)-e_n(a)\log\left(\frac{1-a}2\right)\frac1n-\frac{e_n(a)}n+O\left(\frac1{n^2}\right)$$
Summing these yields $$t_n(a)=\frac{1+a}2\log\left(\frac{1+a}2\right)+\frac{1-a}2\log\left(\frac{1-a}2\right)+e_n(a)\log\left(\frac{1+a}{1-a}\right)+O\left(\frac1{n^2}\right)$$ or $$t_n(a)=\log\alpha(a)+e_n(a)\log\left(\frac{1+a}{1-a}\right)+O\left(\frac1{n^2}\right)$$ which yields the exact equivalent $${n\choose r_n(a)n}\sim\sqrt{\frac2{\pi (1-a^2)n}}\,\alpha(a)^{-n}\beta(a)^{ne_n(a)}$$ where $$\beta(a)=\frac{1-a}{1+a}$$
Now, for every fixed $r$ and $k$, $${n\choose rn+k}\sim{n\choose rn}\left(\frac{1-r}r\right)^k$$ hence $$\sum_{k\geqslant0}{n\choose rn+k}\sim{n\choose rn}\sum_{k\geqslant0}\left(\frac{1-r}r\right)^k={n\choose rn}\frac{r}{2r-1}$$ and $$\sum_{k\geqslant r_n(a)n}{n\choose k}\sim{n\choose r_n(a)n}\frac{1+a}{2a}$$ Finally, $$\sum_{k\in B_n(a)}{n\choose k}=2\sum_{k\geqslant r_n(a)n}{n\choose k}\sim\frac{1}{a}\sqrt{\frac{2(1+a)}{\pi (1-a)n}}\,\alpha(a)^{-n}\beta(a)^{ne_n(a)}$$ hence $$S_n(a)\sim\gamma(a)\,\beta(a)^{ne_n(a)}$$ with $$\gamma(a)=\frac{1}{a}\sqrt{\frac{2(1+a)}{\pi (1-a)}}$$ and where we recall that $$e_n(a)=\left\lceil\frac{1+a}2n\right\rceil-\frac{1+a}2n$$ Due to the oscillations of $e_n(a)$ term, this shows the following:


*

*$S_n(a)$ does not converge, except possibly to $0$ if $a$ is irrational (which disproves your conjecture).

*For every $a$, $\liminf S_n(a)=0$.

*When $a$ is rational, $\limsup S_n(a)=\gamma(a)$. 


For example, if $a=.2=\frac15$, $$\lim S_{5n}(a)=5\sqrt{\frac{3}{\pi}}$$
while $$S_{5n+1}(a)\sim5\sqrt{\frac{3}{\pi}}\left(\frac23\right)^{2n+2/5}\qquad S_{5n+2}(a)\sim5\sqrt{\frac{3}{\pi}}\left(\frac23\right)^{4n+8/5}$$
and $$S_{5n+3}(a)\sim5\sqrt{\frac{3}{\pi}}\left(\frac23\right)^{n+3/5}\qquad S_{5n+4}(a)\sim5\sqrt{\frac{3}{\pi}}\left(\frac23\right)^{3n+12/5}$$
And it might be the case that $\limsup S_n(a)=0$ when $a$ is irrational, depending on $ne_n(a)$ converging to infinity or not.
