I am trying to maximize the function $$f\_s(k)=\frac{1}{2k+s}\sum\_{i=0}^k {2k+s\choose i}2^{-(2k+s)}$$ for both $\{s,k\}\in\mathbf N$, that is, for fixed $s$ what is the value of $k$ that maximizes $f\_s(k)$. I have seen previous discussions in [Sum of 'the first k' binomial coefficients for fixed n](https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n)  and [Lower bound for sum of binomial coefficients?](https://mathoverflow.net/questions/55585/lower-bound-for-sum-of-binomial-coefficients) but I'm not seeing much light. I'd be happy enough with a partial result, for instance a lower bound on the maximum value and the $k$ achieves it, or the range in which the maximum $k$ is to be found.


Anyone any idea? 



Maximize a weighed combinatorial sum
Using Stirling's formula one sees that $\binom{n}{n/2+\ell}$ is about 
$$
\sqrt{\frac{2}{\pi n}} 2^{n} e^{-2\ell^2/n},
$$ 
when $\ell$ is not too large. You can use this to evaluate your sum. 


Let's look at the range when $k$ is of size about $s^2$ which is where the 
maximum will be attained. Writing $n=2k+s = s^2/t$ your sum is approximately 
$$
\frac{1}{\sqrt{2\pi} s^2} \left( t \int\_{\sqrt{t}}^{\infty} e^{-x^2/2} dx\right). 
$$
Choose $t$ such that the term in brackets is maximum -- this is not any nice 
constant so far as I know, it is what it is. It then follows that your 
maximum sum is roughly $C/s^2$ for some constant $C$.


A calculation shows that $t$ above is roughly $1.42$, and that $C$ is about $0.165$. 
Note that then $k$ is about $s^2/2.84$. This is in rough agreement with the $s^2/3$ seen in Meyerowitz's answer; also when $s=100$ the asymptotic $0.165/s^2$ is very close to the $s^{-2.38}$ seen there. 


