TITLE: Sum with binomial coefficients and a square root
QUESTION [16 upvotes]: 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?

REPLY [5 votes]: 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 $S(n)$.