TITLE: An optimization problem involving sum of binomial coefficients upto some value
QUESTION [5 upvotes]: I would like to minimize $f(s, n, \epsilon)$ with respect to $s$ where
$$f(s,n,\epsilon) = \left( 1 + \frac{n}{2^s} \right)\frac{1}{s} \sum_{k=0}^{\lfloor s\epsilon \rfloor} {s \choose k}~.$$
Note that $0 < \epsilon < \frac{1}{2}$ and $n > 0$.
Clearly optimal $s$ is going to be a function of $\epsilon$ and $n$, which might be ugly. However, I think $s^*$ should be close to $\log_2{n}$, based on the intuition from the problem giving rise to this, but I cannot find any rigorous argument for this choice of $s^*$. Any hint or idea is highly appreciated.
The following upper bound might be helpful:
$$ \sum_{k=0}^{\lfloor s\epsilon \rfloor} {s \choose k} \le 2 ^ {H(\epsilon)s } $$
where $H(\epsilon) \equiv -\epsilon \log \epsilon -(1-\epsilon)
\log(1-\epsilon)$; the entropy of a Bernoulli dist. with probability $\epsilon$. 
I don't know of any clean lower bound. Any idea?
Thanks for your time in advance.

REPLY [4 votes]: I assume you have $n\to\infty$. If $\epsilon<\frac12$ and $s\to\infty$ then $$\sum_{k=0}^{\lfloor \epsilon s\rfloor} \binom{s}{k}$$ is very close to a geometric  progression at the big end. Together with Stirling's approximation (ignoring the floor), this gives
$$\sum_{k=0}^{\lfloor \epsilon s\rfloor} \binom{s}{k}
  \approx \frac{1-\epsilon}{1-2\epsilon}\binom{s}{\lfloor \epsilon s\rfloor}
   \approx \frac{1}{\sqrt{2\pi s}}\frac{1-\epsilon}{1-2\epsilon} (\epsilon(1-\epsilon))^{-s-1/2}.
$$
Now you can look for the minimum wrt $s$ by differentiating.  I don't get a closed form but it seems that the minimum is around $s=\log_2(n)+O(1)$.
ADDED: Note that the function is not continuous.  Due to the floor function, it jumps up by a ratio close to $(1-\epsilon)/\epsilon$ as $\epsilon s$ passes an integer.  So there are very many local minima. However, the global minimum ought to be close to the minimum of the function without the floor.