I recently ran into this problem. $X$ follows binomial distribution $B(n,p)$ with $p=\frac1{\sqrt{n}}$. I don't know how to calculate $E\left[\sqrt{X}\right]$. I don't need to find out the exact answer, but I need to find out how it relates to $O\left(\sqrt[4]{n}\right)$.
I'm sorry I didn't give my previous attempts. In fact I am trying to solve for the value of $\sum_{x=1}^{N/2}\tbinom{N/2}{x}(1/\sqrt{N})^x(1-1/\sqrt{N})^{N/2-x}\sqrt{x/n}=\sqrt{1/N}E(\sqrt{x})$. I have tried to try to find this value by solving for $E(x)$. So I get： $$E(\sqrt{x})=\sum_{x=1}^{N/2}{\sqrt{\frac{N}{4x}}\tbinom{N/2-1}{x-1}(1/\sqrt{n})^{x-1}(1-1/\sqrt{n})^{N/2-x}}$$ It is difficult for me to further calculate the value of this equation, so I tried to discover the relationship between this value and $O\left(\sqrt[4]{n}\right)$. I let $x=N^{1/4}$. I get: $${\sqrt{\frac{N}{4N^{1/4}}}\tbinom{N/2-1}{N^{1/4}-1}(1/\sqrt{n})^{N^{1/4}-1}(1-1/\sqrt{n})^{N/2-N^{1/4}}}$$ I try to use Stirling's approximation to calculate $\tbinom{N/2-1}{N^{1/4}-1}$, but here is still no way to simplify $(1-1/\sqrt{n})^{N/2-N^{1/4}}$ with $\tbinom{N/2-1}{N^{1/4}-1}$.
So I was wondering if this expectation had been solved before, but I didn't find anything about it.
If you just need to show that $\mathbb{E}[\sqrt{X}] = O(n^{1/4})$, then since $\mathbb{E}[X]-\mathbb{E}[\sqrt{X}]^2 = \text{Var}[\sqrt{X}] \ge 0$, we have that $\mathbb{E}[\sqrt{X}] \le \sqrt{\mathbb{E}[X]} = \sqrt{n \cdot \tfrac{1}{\sqrt{n}}} = n^{1/4}$.
This is enough to show that $\sqrt{\mathbb{E}[X]} = O(n^{1/4})$, but if you also want a crude lower bound, that can be done as well:
Since the median of a $B(n,p)$ distribution is either $\lfloor np \rfloor$ or $\lceil np \rceil$, we have that $\mathbb{P}[X \ge \lfloor n^{1/2} \rfloor] \ge \dfrac{1}{2}$. Hence, $$\begin{align*}\mathbb{E}[\sqrt{X}] &= \sum_{k = 0}^{n}\mathbb{P}[X = k]\sqrt{k} \\ &= \sum_{k = 0}^{\lfloor n^{1/2} \rfloor - 1}\mathbb{P}[X = k]\sqrt{k} + \sum_{\lfloor n^{1/2} \rfloor}^{n}\mathbb{P}[X = k]\sqrt{k} \\ &\ge \sum_{k = 0}^{\lfloor n^{1/2} \rfloor - 1}\mathbb{P}[X = k] \cdot 0 + \sum_{\lfloor n^{1/2} \rfloor}^{n}\mathbb{P}[X = k] \cdot \sqrt{\lfloor n^{1/2} \rfloor} \\ &= \mathbb{P}[X \ge \lfloor n^{1/2} \rfloor] \cdot \sqrt{\lfloor n^{1/2} \rfloor} \\ &\ge \dfrac{1}{2}\sqrt{\lfloor n^{1/2} \rfloor}\end{align*}$$
Therefore, $\dfrac{1}{2}\sqrt{\lfloor n^{1/2} \rfloor} \le \mathbb{E}[\sqrt{X}] \le n^{1/4}$.
We can probably get a stronger lower bound on $\mathbb{E}[\sqrt{X}]$ by using tail bounds for the binomial distribution.