In a paper that I am reading, the author states some simple asymptotic behavior about a sequence, in which I got a different result. I would like to know where my analysis went wrong.
Let $N=2^n, n\geq 0$, $0\leq K\leq N$ where $n, K$ are all nonnegative integers. Let the integer parameter $r$ satisfying the equation
$$\sum_{k=r}^{n}\binom{n}{k}\leq K <\sum_{k=r-1}^n \binom{n}{k}$$
be denoted as $r(N)$. The author claims that as $n$ goes to infinity (so that $N$ goes to infinity through the powers of two), and $K$ is decided to be $K=\lfloor NR\rfloor$ where $R$ is some fixed value in $(0, 1)$, then the ratio $\frac{r(N)}{n}=\frac{r(N)}{\log_2 N}$ must go to $\frac{1}{2}$, implying that the value of $R$ is not important.
If the author's claim is true, then when $n$ goes to $\infty$, then the left hand side of the inequality $\sum_{k=r}^n\binom{n}{k}$ goes to $$\sum_{k=\frac{n}{2}}^{n}\binom{n}{k}\to 2^{n-1}=\frac{N}{2}$$ which makes no sense when $R\in (0, \frac{1}{2})$ so that $K=\lfloor NR \rfloor<\frac{N}{2}$. Where did my analysis go wrong?
Let $X$ be the sum of $n$ fair Bernoulli variables, so $X$ is Binomial, $B(n,\frac12)$. Let $Z=\frac{1}{n}X$.
Then $$ \sum_{k=r}^{n} \binom{n}{k} = 2^n P(X\ge r) = N \,P(Z\ge r/n) \tag1$$
and
$$ P(Z\ge r/n) \le \frac{K}{N} < P(Z\ge r/n -1/n) \tag2$$
$$ P(Z\ge r/n) \le R -\frac{\delta}{2^n} < P(Z\ge r/n - 1/n) \tag3$$
for some $0\le \delta<1$. As $n \to \infty$, the center term in $(3)$ tends to $R$, and in the limit $R=P(Z\ge r/n) $.
Now, $Z$ has mean $\mu=\frac12$ and its variance $\sigma^2= \frac{1}{4 \sqrt{n}}$ vanishes with $n$. Hence, if $0<R<1$ (fixed value, strictly distinct from $\{0,1\}$) the value $r/n$ must tend to $\frac{1}{2}$. Because for any $a < \frac12$, $P(Z\ge a) \to 1$ and for any $a > \frac12$ , $P(Z\ge a) \to 0$.
Hence, the paper is right.
The problem with your reasoning, in your last paragraph, is that you seem to assume that $\frac{r}{n} \to \frac12 \implies r \to \frac{n}{2}$ (and hence $\sum_{k=r}^{n} (\cdot) \to \sum_{k=n/2}^{n} (\cdot) $ ). That is, of course, wrong. In particular, in our case the CLT says that the growth rate will be $r= \frac{n}{2} + \alpha \sqrt{n} + O(1)$ where $\alpha = \alpha(R)$