Sum with binomial coefficients and a square root I encountered this sum from working on an integral:
$$\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}\sqrt{k}$$
I don't think it can be written as a hypergeometric function, because of this square root.
Does this sum have a closed form?
 A: Here is some more information which also supports the assumption from Noam D.Elkies, that it is rather unlikely to find a simpler expression for the sum

$$S(n) = \sum_{k=0}^{n}\binom{n}{k}(-1)^k\sqrt{k}$$

We could try to simplify the sum using generating functions and transform the power series

$$\sum_{n\ge 0}S(n)z^n  = \sum_{n\ge 0}\left(\sum_{k=0}^{n}\binom{n}{k}(-1)^k\sqrt{k}\right)z^n$$

with

Euler's series transformation formula:
Given a function $f(z)=\sum_{n\ge0}a_nz^n$ analytical on the unit disk, the following representation is valid:
  \begin{align}
f(z)&=\sum_{n\ge0}a_nz^n\\
\frac{1}{1-z}f\left(\frac{z}{1-z}\right)&=\sum_{n\ge0}\left(\sum_{k=0}^{n}\binom{n}{k}a_k\right)z^n\tag{1}
\end{align}

This transformation formula together with a proof using Cauchy's integral formula can be found in Harmonic Number Identities Via Euler's transform from Boyadzhiev ($2009$).
We observe that $a_n = (-1)^n\sqrt{n}$ and so we get the Polylogarithm $Li_{-\frac{1}{2}}(-z)$

$$f(z)=Li_{-\frac{1}{2}}(-z)=\sum_{n\ge1}\frac{(-z)^n}{n^{-\frac{1}{2}}}=\sum_{n\ge1}(-1)^n\sqrt{n}z^n$$

According to $(1)$ we see that $S(n)$ is the coefficient of $z^n$ from: 

$$S(n)=\sum_{k=0}^{n}\binom{n}{k}(-1)^k\sqrt{k} = [z^n]\frac{1}{1+z}Li_{-\frac{1}{2}}\left(\frac{-z}{1+z}\right)$$

It' don't seem plausible (for me), that this polylogarithm could be properly transformed in order to retrieve a simpler expression for $S(n)$.
A: A closed form for
$$
S(n) := \sum_{k=0}^n \binom{n}{k} (-1)^{k} \sqrt{k}
$$
is indeed unlikely; at any rate it surely can't be a finite
hypergeometric sum, because the terms with $k$ squarefree in 
$n/4 < k \leq n$ contribute independent square roots, 
and there are about $\frac34 n / \zeta(2)$ of these
so the degree of $S(n)$ as an algebraic number grows
exponentially in $n$.
One can, however, obtain a definite integral formula for $S(n)$
(indeed the OP's question suggested he or she started with such an integral):
$$
S(n) = -\frac1{\sqrt\pi} \int_0^\infty \left(1-e^{-x^2}\right)^n \frac{dx}{x^2}
$$
for all $n>0$.  (To obtain this formula, start with the familiar
$\int_0^\infty e^{-ax^2} dx = \frac12 \sqrt{\pi/a}$, integrate from
$a=0$ to $a=k$ and divide by $\sqrt\pi$ to obtain
$$
\sqrt{k} = 
 \frac1{\sqrt\pi} \int_0^\infty \left(1-e^{-kx^2}\right) \frac{dx}{x^2},
$$
and then multiply by $(-1)^k \binom{n}{k}$ and sum from $k=0$ to $k=n$.)
For example, it follows from the integral that
$$
-1 = S(1) < S(2) < S(3) < S(4) < \cdots \rightarrow 0
$$
but the convergence to zero is very slow, with $S(n)$ asymptotic to
$-(\pi \log n)^{-1/2}$ for large $n$; for example
$S(10^4)$ is still $-.1814\ldots\;$.$\ $  [If numerical calculation
suggests that $S(n)$ behaves more irregularly than this,
beware of massive cancellation in the defining sum, whose
largest terms are proportional to $\pm 2^n$; I used several
thousand digits of precision to compute the numerical value of $S(10^4)$.
At any rate it is clear that $S(n)<0$ for $n>0$,
because it is $(-1)^n$ times an $n$-th finite difference of
a function whose $n$-th derivative has sign $(-1)^{n+1}$ everywhere.]
