Convergence of probability of throwing at least $\lceil\frac{n}2\rceil$ heads in $n$ fair coin tosses I recently provided an answer to a question that asked about the probability of throwing at least some number of heads in a number of fair coin flips.  Extending my thinking, I came to a conclusion that if $n$ is odd, then the probability of throwing at least $\lceil\frac{n}2\rceil$ heads is always $0.5$ (due to symmetry:  $\binom{n}{r}=\binom{n}{n-r})$.  For even $n$, the probability approaches $0.5$.  I want to know if it is correct to say
$$\lim_{n\to\infty}\frac{\sum_{r=\lceil\frac{n}{2}\rceil}^n\binom{n}{r}}{\sum_{r=0}^{n}\binom{n}{r}}=0.5$$ 
I understand that if we defined $n\in\mathbb{R}$, then we would simply have the limit
$$\lim_{n\to\infty}\frac{\lceil\frac{n}2\rceil}{n}$$
(which I am not sure how to evaluate), but since $n\in\mathbb{Z^+}$, I don't think that it is entirely fair.
Basically, since for odd $n$ the value is a constant $0.5$, and for even $n$ the value approaches $0.5$, do we say that the series converges, or do we say that the series diverges?
 A: If $n=2m$, the probability of getting at least $m$ heads is by symmetry 
$$
p_m=\frac{1}{2}-\frac{\binom{2m}{m}}{2^{2m}}
$$
Let $u_m=\frac{\binom{2m}{m}}{2^{2m}}$. It is easy to check that 
$\frac{u_{m+1}}{u_m}=1-\frac{1}{2m+2}$, so $u_m=u_1\prod_{k=1}^{m}\big(1-\frac{1}{2k+2}\big)$ whence $\ln(u_m)=\ln(u_1)+\sum_{k=1}^{m}\ln\big(1-\frac{1}{2k+2}\big)$. Now $\sum_{k\geq 1} \ln\big(1-\frac{1}{2k+2}\big)$ behaves as $\sum_{k\geq 1}-\frac{1}{2k+2}$, i.e. it diverges to $-\infty$, so that $u_n\to 0$ as wished.
A: As you can see in this post, there are fully self-contained ways of answering your question that don't mention probability. However, I believe it is more natural - and instructive - to answer your question using one of the most fundamental probabilistic tools: the central limit theorem.
The central limit theorem is very general, but when applied to the case at hand involving coin flips it boils down to a very explicit identity:
$$
\lim_{n\to\infty}\sum_{k=k^-}^{k^+} \binom{n}{k}\cdot \frac{1}{2^n}=\frac{1}{\sqrt{2\pi}}\int_{x-\epsilon}^{x+\epsilon} e^{-t^2/2}\ dt,
$$
where ${k^{-}=\Bigl\lfloor \frac{n}{2}+(x-\epsilon)\frac{\sqrt{n}}{2}\Bigr\rfloor}$ and ${k^{+}=\Bigl\lceil \frac{n}{2}+(x+\epsilon)\frac{\sqrt{n}}{2}\Bigr\rceil}$.
In particular, if we take $x=\epsilon=0$ and $n=2m$ then it says
$$
\lim_{m\to\infty}\binom{2m}{m}\cdot \frac{1}{2^{2m}}=\frac{1}{\sqrt{2\pi}}\int_{0}^{0} e^{-t^2/2}\ dt,
$$
and since the integral on the right is obviously $0$, it answers your question.
Regarding your follow-up question, the sequence still converges to $\tfrac{1}{2}$ even though the odd values exactly equal $\tfrac{1}{2}$ while the even values are close to, but not exactly $\tfrac{1}{2}$. This is for the same reason that a constant sequence $a_n=\tfrac{1}{2}$ also converges to $\tfrac{1}{2}$ (although in the most boring way possible).
