Is the maximum value $a$ such that the sum of the middle binomial coefficients
$$2^{-n} \sum_{k=(n/2)-\lfloor n^a \rfloor}^{(n/2)+\lfloor n^a \rfloor} C_k^n$$
goes to zero known?
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1I believe there's no maximum, but that the supremum is $a=\frac12$; I haven't looked closely but this should follow almost trivially from the Central Limit theorem. – Steven Stadnicki Sep 03 '13 at 05:30
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Introducing binomial $(n,\frac12)$ random variables $X_n$, one sees that the sum in the question is $$ p_n^{(a)}=P[|X_n-\tfrac12n|\leqslant n^a]. $$ In particular $E[X_n]=\frac12n$ and $\mathrm{var}(X_n)=\frac14n$ hence Bienaymé-Chebyshev inequality yields $$p_n^{(a)}\leqslant\frac{n^{2a}}{\mathrm{var}(X_n)}=4n^{2a-1},$$ which shows that $p_n^{(a)}\to0$ for every $a\lt\frac12$.
On the other hand, $p_n^{(1/2)}=P[|X_n-\frac12n|\leqslant2\sqrt{\mathrm{var}(X_n)}]$ hence, by the central limit theorem, $p_n^{(1/2)}\to P[|Z|\leqslant2]$, where $Z$ is standard normal. Since $P[|Z|\leqslant2]\ne0$, $p_n^{(1/2)}$ does not converge to $0$.
Finally:
The set of $a$ such that $p_n^{(a)}\to0$ is $(-\infty,\frac12)$. Note that this set has no "maximal value", only the supremum $\frac12$.
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