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

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.
