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
 A: 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.
A: 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 $\sum_{n=1}^\infty (-1)^{n+1}\sqrt{n}$
A: 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.
A: 
I thought it might be instructive to present a brute force approach.  To that end we proceed.

Let $S_n=\sum_{k=1}^n (-1)^{k-1}\sqrt{k}$ be the sequence of interest.  Then, we can write the even and odd terms, respectively by
$$\begin{align}
S_{2n}&=\sum_{k=1}^n \left(\sqrt{2k-1}-\sqrt{2k}\right)\\\\
S_{2n+1}&=1+\sum_{k=1}^n \left(\sqrt{2k+1}-\sqrt{2k}\right)
\end{align}$$
Then, the Cesaro Sum is given by
$$\begin{align}
\frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}&=\frac{1+\sum_{n=1}^N\left(1+ \sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\right)}{2N+1} \tag 1\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{n=1}^N\sum_{k=1}^n \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 2\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N\sum_{n=k}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)\tag 3\\\\
&=\frac{N+1}{2N+1}+\frac{1}{2N+1}\sum_{k=1}^N(N+1-k)\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 4\\\\
&=\frac{N+1}{2N+1}+\frac{N+1}{2N+1}\sum_{k=1}^N \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) \tag 5\\\\
&-\frac{1}{2N+1}\sum_{k=1}^N k\left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)
\end{align}$$


NOTES:
In going from $(1)$ to $(2)$, we simply carried our the trivial sum $\sum_{k=1}^n (1)=N$.
In going from $(2)$ to $(3)$, we interchanged the order of summation.
In going from $(3)$ to $(4)$, we evaluated the inner sum.
In going from $(4)$ to $(5)$, we split the expression into the sum of three terms.


As $N\to \infty$, the first term in $(5)$ approaches $\frac12$ while the second term approaches $\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right)$.  The third term in $(5)$ can be shown to approach $0$ since $\sum_{k=1}^N k \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) =O\left(N^{1/2}\right)$.
Therefore, we find that 
$$\lim_{N\to \infty}\frac{\sum_{n=0}^N (S_{2n}+S_{2n+1})}{2N+1}=\frac12+\frac12 \sum_{k=1}^\infty \left(\sqrt{2k+1}-2\sqrt{2k}+\sqrt{2k-1}\right) $$
A: Hopefully I'm not copying Gottfried Helms' Euler sum, but the Euler summation formula allows you to sum this divergent series.
$$\sum_{k=1}^\infty(-1)^{k+1}\sqrt k=\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$
This converges extremely quickly to the Riemann zeta function,
$$\zeta(-1/2)=\frac1{1-\sqrt8}\sum_{n=0}^\infty\frac1{2^n}\sum_{k=0}^n\binom nk(-1)^k\sqrt{k+1}$$
