Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable?

How about an Abel sum? $$ \sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}\;x^n = -\mathrm{Li}_{-1/2}(-x) $$ for $|x|<1$ and converges as $x \to 1^-$ to the value $$ -\mathrm{Li}_{-1/2}(-1) \approx 0.3801048 $$ So we call that value the Abel sum of the divergent series $\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}$

Let us show that $\sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k}$ is Cesaro summable. Once we establish this, then this is also Abel summable and the Cesaro sum is equal the Abel sum, which is

$$ \sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k} = -\operatorname{Li}_{-1/2}(-1) = (1 - 2^{3/2})\zeta(-1/2). $$

To this end, let $s_n = \sum_{k=1}^{n} (-1)^{k-1}\sqrt{k}$ and notice that $s_n = \mathcal{O}(\sqrt{n})$. This can be easily checked by grouping two successive terms and applying the mean value theorem. Thus it suffices to prove that

$$ \frac{s_1 + \cdots + s_{2n+1}}{2n+1} $$

converges. Now the trick is to consider

$$ s_{2n} + s_{2n+1} = 1 + \sum_{k=1}^{n} (\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k}). $$

Using Taylor series, it is not hard to check that

$$\sqrt{2k-1} + \sqrt{2k+1} - 2\sqrt{2k} = \mathcal{O}(k^{-3/2}). $$

Thus $s_{2n} + s_{2n+1}$ converges as $n \to \infty$, and the claim follows from Cearso-Stolz theorem.

This sum can be done with some form of zeta function regularization. For $\Re s >1$, define:

$$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s) $$

Then, by analytic continuation, we can calculate:

$$\sum_{n=1}^\infty (-1)^{n-1} \sqrt{n} \to \eta\left(-\frac{1}{2}\right) = (1-2\sqrt{2})\zeta\left(-\frac{1}{2}\right) \approx .3801048$$

This is equal to the Abel sum $-\operatorname{Li}_{-\frac{1}{2}} (-1)$ from GEdgar's answer.