TITLE: Sum of square roots of binomial coefficients
QUESTION [7 upvotes]: Question: let 
\begin{equation}
c_n=\sum_{i=0}^n \sqrt{\binom{n}{i}}.
\end{equation}
How does $c_n$ grow with $n$?
My conjecture is that $c_n=\Theta(2^{0.5n}n^{0.25})$.This is because 
\begin{equation}
 0.5^{0.5n} c_n = \sum_{i=0}^n \sqrt{\binom{n}{i}0.5^n}=\sum_{i=0}^n\sqrt{Pr(X=i)},
\end{equation}
for $X\sim Bin(n,0.5)$. The binomial distribution $Bin(n,0.5)$ is approximately the normal distribution $N(0.5n, 0.25n)$. Also, if $Y\sim N(0.5n, 0.25n)$, it is not hard to see that 
\begin{equation}
\int \sqrt{f(y)} dy =\Theta(n^{0.25}). 
\end{equation}
Therefore I believe that it is also true that $0.5^{0.5n} c_n=\Theta(n^{0.25})$. I played with matlab to get some evidence and found that 
\begin{equation}
\frac{0.5^{0.5n} c_n}{n^{0.25}} \approx \frac{\pi}{2},
\end{equation}
for $n=100,1000,10000,100000,1000000$. So the conjecture should be true. So anyone can prove it?
(It turns out that $\frac{0.5^{0.5n} c_n}{n^{0.25}} \approx (2\pi)^{0.25}$ instead of $\frac{\pi}{2}$).

REPLY [9 votes]: For smallish $k$, we have
$$ \binom{n}{n/2+k} \approx \binom{n}{n/2} \exp(-2k^2/n). $$
So
$$\sum\sqrt{\binom{n}{n/2+k}}
  \approx  \sqrt{\binom{n}{n/2}}  \int_{-\infty}^\infty e^{-k^2/n}\,dk
  \approx 2^{n/2} (2\pi n)^{1/4}.$$
Of course I did some hand-waving here, but all of this is rigorously justifiable.  Use the Euler-Maclaurin theorm to justify replacing the sum by an integral.