Norbert's answer is the simplest with the fewest prerequisites. That being the case, here is a nuke for an ant hill using the Euler-Maclaurin Sum Formula.
Note that $ \begin{align} \sum_{k=1}^{2n}(-1)^k\sqrt{k} &=2\sum_{k=1}^{n}\sqrt{2k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\\ &=\sqrt{8}\sum_{k=1}^{n}\sqrt{k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\tag{1} \end{align} $ The Euler-Maclaurin Sum Formula says that $ \sum_{k=1}^n\sqrt{k}=\frac23n^{3/2}+\frac12n^{1/2}+C+\frac{1}{24}n^{-1/2}+O\left(n^{-3/2}\right)\tag{2} $ For $\mathrm{Re}(z)>-1$, $ \zeta(z)=\lim_{n\to\infty}\sum_{k=1}^nk^{-z}\;-\;\left(\frac{1}{1-z}n^{1-z}+\frac12n^{-z}\right)\tag{3} $ Applying $(3)$ to $(2)$ yields $C=\zeta(-\frac12)=-\frac{1}{4\pi}\zeta(\frac32)\,\dot{=}-0.207886224977354566$.
Applying $(2)$ to $(1)$ yields $ \begin{align} \sum_{k=1}^{2n}(-1)^k\sqrt{k} &=\sqrt{8}\sum_{k=1}^{n}\sqrt{k}\;-\;\sum_{k=1}^{2n}\sqrt{k}\\ &=\sqrt{8}\left(\frac23n^{3/2}+\frac12n^{1/2}+C+\frac{1}{24}n^{-1/2}\right)\\ &-\left(\sqrt{8}\frac23n^{3/2}+\sqrt2{}\frac12n^{1/2}+C+\frac{1}{\sqrt{2}}\frac{1}{24}n^{-1/2}\right)+O\left(n^{-3/2}\right)\\ &=\frac{1}{\sqrt{2}}n^{1/2}+(\sqrt{8}-1)C+\frac{3}{\sqrt{2}}\frac{1}{24}n^{-1/2}+O\left(n^{-3/2}\right)\tag{4} \end{align} $ Thus, for even $n$, we get $ \sum_{k=1}^{n}(-1)^k\sqrt{k}=\frac12n^{1/2}+\frac18n^{-1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\tag{5} $ For odd $n$, we get $ \begin{align} \sum_{k=1}^{n}(-1)^k\sqrt{k} &=\sum_{k=1}^{n-1}(-1)^k\sqrt{k}\;-\;\sqrt{n}\\ &=\color{red}{\frac12(n-1)^{1/2}}+\color{green}{\frac18(n-1)^{-1/2}}-n^{1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\\ &=\color{red}{\frac12n^{1/2}-\frac14n^{-1/2}}+\color{green}{\frac18n^{-1/2}}-n^{1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\\ &=-\frac12n^{1/2}-\frac18n^{-1/2}+\frac{7C}{\sqrt{8}+1}+O\left(n^{-3/2}\right)\tag{6} \end{align} $ using the $\color{red}{\text{Binomial}}$ $\color{green}{\text{Theorem}}$ to expand the powers of $(n-1)$ to $O\left(n^{-3/2}\right)$.
Combining $(5)$ and $(6)$, we get $ \sum_{k=1}^{n}(-1)^k\sqrt{k}=(-1)^n\left(\frac12n^{1/2}+\frac18n^{-1/2}\right)-\frac{7}{\sqrt{8}+1}\frac{\zeta(\frac32)}{4\pi}+O\left(n^{-3/2}\right)\tag{7} $ We've used more powerful machinery, but we've also gotten a more precise answer.