Partial sum of binomial coefficients For a given $n$, find the minimal $r$ such that.
$$2^r\sum_{i=0}^r {n \choose i} \ge 2^n$$
What can we say about $\frac{r}{n}$ as $n \to \infty$?
I wrote a small script and it looks like there's a limit (around 0.22709), although the expression is not monotonically decreasing, here's the
graph.
I wonder if there's anything analytical to say about this question, to prove if a limit exists, or to bound this value.
For context, this question came up when I was looking for particular coverings of $\Bbb Z_2^n$ with hamming spheres, specifically $2^r$ spheres with radius r.
 A: I'll be a little sloppy below; the goal will only be to figure out the constant. This argument could be made rigorous with more careful attention to bounds. Below $\approx$ means that two expressions are within a polynomial-of-$n$ factor of each other.
Write $a_r = \sum_{i=0}^r {n \choose i}$. We have the crude bounds ${n \choose r} \le a_r \le r  {n \choose r}$, so $a_r \approx {n \choose r}$. Writing $r = pn$, Stirling's approximation then gives
$$\begin{align*} {n \choose pn} &\approx \frac{ \left( \frac{n}{e} \right)^n}{ \left( \frac{pn}{e} \right)^{pn} \left( \frac{(1-p)n}{e} \right)^{(1-p)n} } \\
 &= \frac{n^n}{(pn)^{pn} ((1-p)n)^{(1-p)n} } \\
 &= \frac{1}{p^{pn} (1-p)^{(1-p)n}} \\
 &= 2^{n H_2(p)} \end{align*}$$
where $H_2(p) = - p \log_2 p - (1 - p) \log_2 (1 - p)$ is the binary entropy; this is the entropy approximation to the binomial coefficients. So we want to find $p$ such that
$$2^r {n \choose r} \approx 2^{n(p + H_2(p))} \approx 2^n$$
which means the value of $p$ we want is the smallest positive real solution to
$$\boxed{p + H_2(p) = 1}.$$
WolframAlpha can plot this function but for whatever reason can't solve this equation even numerically.

In any case the plot clearly shows a solution around $p = 0.22 \dots $ as you've found, and since the function is so well-behaved Newton's method can compute $p$ quickly to high accuracy. Here the Newton's method iteration is
$$p_{n+1} = p_n - \frac{p_n + H(p_n) - 1}{1 - \log_2 \frac{p_n}{1-p_n}}$$
and we can take, say, $p_0 = 0.2$ as our initial guess. After three iterations we get $\boxed{ p_3 = 0.2270921952 \dots }$ and these digits are stable after further iteration (at least up to the accuracy Google Sheets uses by default).
Generally, understanding the asymptotics of $a_r$ amounts to understanding the probability that a binomial distribution is far from its mean. This is the subject of large deviations inequalities, and in particular Cramer's theorem can be applied here to produce a bound that will be a more rigorous version of the above.
