Does $\sum \frac {(-1)^k}{\sqrt{k}}$ have a closed form? Does $\sum \frac {(-1)^k}{\sqrt{k}}$ have a closed form? is there a general known formula for $\sum \frac {(-1)^k}{\sqrt [p]{k}}$ ? I have seen many results relating to $\sum \frac {(-1)^k}{k^p}$ but do not recall coming across results relating to $\sum \frac {(-1)^k}{\sqrt [p]{k}}$ , is there a name for such series?
I am not assuming classical convergence although it seems obvious.
 A: The Dirichlet Eta Function is defined as:
$$\eta(s) = \sum_{n= 1}^\infty \frac{(-1)^{n-1}}{n^s}$$
You may notice that if we assign $s:= \frac{1}{p}$, then we have that:
$$-\eta(1/p) = (-1)\times\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n^{1/p}} = \sum_{k = 1}^\infty \frac{(-1)^k}{\sqrt[p]{n}}$$
So, if $\eta(s)$ is known at specific values $\frac{1}{p}$, you have your answer.  Unfortunately, it appears specific values are only really known for integer values (so $\eta(2),\eta(3)$,etc), not for fractional amounts.  I hope this gives you more to look for (regarding the specific name of the series).
A: The analytic continuation of the Riemann $\zeta$ function through the $\eta$ function gives the identity:
$$ \zeta(s) = \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \left(1-\frac{2}{2^s}\right)^{-1} \eta(s) \tag{1}$$
for any $s\in\mathbb{C}$ such that $\text{Re}(s)>0$. In particular:
$$ \sum_{k\geq 1}\frac{(-1)^k}{\sqrt{k}} = -\eta\left(\frac{1}{2}\right) = (\sqrt{2}-1)\,\zeta\left(\frac{1}{2}\right).\tag{2} $$
The (inverse) Laplace transform provides accurate numerical approximations:
$$ \sum_{k\geq 1}\frac{(-1)^k}{\sqrt{k}} = \int_{0}^{+\infty}\sum_{k\geq 1}\frac{(-1)^k e^{-ks}}{\sqrt{\pi s}}\,ds = -\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+\exp(u^2)}\tag{3}$$
and since $\frac{1}{1+\exp(u^2)}\approx \left(\frac{1}{2}+\frac{u^2}{4}\right)e^{-u^2}$,
$$ \sum_{k\geq 1}\frac{(-1)^k}{\sqrt{k}} \approx \color{blue}{-\frac{5}{8}}.\tag{4}$$
