TITLE: How to evaluate $\sum\limits_{k=0}^{n} \sqrt{\binom{n}{k}} $
QUESTION [8 upvotes]: Can we find $$ \sum_{k=0}^{n} \sqrt{\binom{n}{k}} \quad$$
This problem asked me my friend about a year ago, but I didn't know how to attack problem. Now, I am interesting in solution. Any suggestion?

REPLY [4 votes]: For any positive power $p$, $\displaystyle \sum\limits_{k=1}^{n} \binom{n}{k}^p \sim \dfrac{2^{np}}{\sqrt{p}}\cdot\left(\frac{n\pi}{2}\right)^{(1-p)/2} $,
Making, $n \mapsto 2n$,
We have, $\displaystyle \dfrac{\binom{2n}{n-k}}{\binom{2n}{n}} = \dfrac{n}{n+k}\prod\limits_{j=1}^{k-1}\dfrac{1-\frac{j}{n}}{1+\frac{j}{n}} = \dfrac{n}{n+k}e^{\sum\limits_{j=1}^{k-1}\log\frac{1-j/n}{1+j/n}}$
and form taylor expansion at $t=0$, $\log \dfrac{1-t}{1+t} = -2t - \dfrac{2x^3}{3}+O(t^5)$,
so, $0 < -2t - \log\dfrac{1-t}{1+t} = O(t^3) $ for, $0 < t < 1/2$.
It sufices to look at the central $n^{2/3}$-terms, 
since, $\displaystyle \dfrac{\binom{2n}{n-k}}{\binom{2n}{n}} = \dfrac{n}{n+k}\large{e^{\sum\limits_{j=1}^{k-1}\log\frac{1-j/n}{1+j/n}}}$ $ = \begin{cases}(1+O(n^{-1/3}))e^{-k^2/n} & \textrm{ for } k\le n^{2/3} \\ O(e^{-n^{1/3}})  & \textrm{ for } k>n^{2/3}\end{cases}$
Hence, $\displaystyle \sum_{k=1}^{n-1} \dfrac{\binom{2n}{n-k}^p}{\binom{2n}{n}^p} = (1+O(n^{-1/3}))\sum\limits_{k=1}^{n^{2/3}} e^{-pk^2/n} + o(1) = (1+O(n^{-1/3}))\int_1^{n^{2/3}} e^{-pt^2/n}\,dt + O(1)$
$$\sim \frac{1}{2}\sqrt{\dfrac{n\pi}{p}}$$
Using Stirling Approximation we have, $\displaystyle \binom{2n}{n} \sim \dfrac{4^{n}}{\sqrt{n\pi}}$
Hence, $\displaystyle \sum\limits_{k=1}^{n} \binom{n}{k}^p \sim \dfrac{2^{np}}{\sqrt{p}}\cdot\left(\frac{n\pi}{2}\right)^{(1-p)/2} $