Let $X\sim B(n,p)$ denote a binomial random variable. Is there any approximation available for the quantity $E(\sqrt{X})$? Clearly Jensen's inequality holds, but rudimentary tooling around with Maple hasn't turned up anything more substantial.



Expectation of square root of binomial r.v.
$\newcommand{\E}{\mathbf{E}}$
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If we use Taylor expansion (as Anthony suggested) for $\sqrt{x}$ around 1, we get:
$$\sqrt{x}\approx 1 + \frac{x-1}{2} - \frac{(x-1)^2}{8} .$$


We can use this to get an approximation of
$$\E(\sqrt{X})\approx 1-\frac{\var(X)}{8} ,$$
which should be valid for any RV concentrated around an expectation of 1. Equivalently,
$$\E(\sqrt{X})\approx \sqrt{\E(X)}\bigg(1-\frac{\var(X)}{8\E(X)^2}\bigg) ,$$
for any RV concentrated around its mean.


As you noted, we can use Jensen inequality to get $\E(\sqrt{X})\le \sqrt{\E(X)}$ for any nonnegative RV. We can tweak the Taylor expansion to get a lower bound, by noticing that
$$ 1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} \ .$$
Hence, we get
$$\sqrt{\E(X)}\bigg(1-\frac{\var(X)}{2\E(X)^2}\bigg) \le \E(\sqrt{X}) ,$$
for any nonnegative RV.


In the case of $X\sim Bin(n,p)$ we get
$$ \sqrt{np}-\frac{1-p}{2\sqrt{np}} \le \E(\sqrt{X})\le \sqrt{np} .$$


