Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable?  I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} + 
 \frac{1}{\sqrt{2}} + \dots +  \frac{1}{\sqrt{n}} \right)
\approx  \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity.  Does any more creative summation method work?

The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy

the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series  $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...

Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK

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.
