Determine Convergence or Divergence of Summation $(-1)^n \sqrt n\over \ln n$ Determine whether the series is conditionally convergent, absolutely convergent, or divergent.

$$\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}$$

The absolute value of this sum is divergent by the divergence test, and it's inconclusive by the ratio and root tests, and the alternating series test doesn't apply because the sequence isn't decreasing. I think all of this only tells me that my conclusions are inconclusive. Is this correct? What can I do to actually determine if this series is absolutely convergent, conditionally convergent, or divergent?
Note that
$$
\lim_{n\to\infty}\frac{\sqrt{n}}{\ln(n)}\stackrel{L}{=}\lim_{n\to\infty}\frac{\frac{1}{2\sqrt{n}}}{\frac{1}{n}}
=\frac{1}{2}\lim_{n\to\infty}\frac{n}{\sqrt{n}}
=\frac{1}{2}\lim_{n\to\infty}\sqrt{n}=\infty
$$
This implies that your sum diverges.
This is not a complete solution, just an observation.

We can rewrite the series as:

$$\sum_{n=2}^{\infty}b_n:=\sum_{n=2}^{\infty}\frac{(-1)^n\sqrt{n}}{\ln(n)}=\sum_{k=1}^{\infty}\left(\frac{\sqrt{2k}}{\ln(2k)}-\frac{\sqrt{2k+1}}{\ln(2k+1)}\right):=\sum_{k=1}^{\infty}a_k$$

It is shown in the first answer that general term $b_n$ is going to $\infty$ when $n\to\infty$:
$$\lim_{n\to\infty}b_n\to \infty.$$

But we should not jump to conclusion based on this fact.

Here we can show that the general term $a_k$ of the transformed series is going to 0 when $k\to\infty$:

$$\lim_{k\to\infty}a_k=\frac{2+\log 1-\log{(2k)}}{2\sqrt{2k}(\log 1-\log{(2k)})^2}\to0.$$

Thus for alternating series $\sum_{n=2}^{\infty}(-1)^n b_n$ it is better to look at the transformed one: $\sum_{k=1}^{\infty}(b_{2k}-b_{2k+1}):=\sum_{k=1}^{\infty}a_k$.

EDIT:
Here is an example that behaves similar to the series in OP.

$$S=\sum_{n=2}^{\infty}b_n=\sum_{n=2}^{\infty}\left((-1)^{n+1}[n/2]+\frac{1}{n^2}\right)\tag{5}$$

where $[n/2]$ is the largest integer that is smaller or equal to $n/2$.

Thus $[(2k)/2]=k$ and $[(2k+1)/2]=k$, for $k\in \mathbb{N}$.

We notice that $|b_n|\to [n/2] \to\infty$ when $n\to\infty$.

On the other hand we have
$$b_{2k}+b_{2k+1}=\left((-1)^{2k+1}[(2k)/2]+\frac{1}{(2k)^2}\right)+\left((-1)^{2k+2}[(2k+1)/2]+\frac{1}{(2k+1)^2}\right)$$
$$=\left(-k+\frac{1}{(2k)^2}\right)+\left(k+\frac{1}{(2k+1)^2}\right)=\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}:=a_k$$

Thus we may rewrite $S$ by grouping $b_{2k}$ and $b_{2k+1}$ as:

$$S=\sum_{k=1}^{\infty}a_k=\sum_{k=1}^{\infty}\left(\frac{1}{(2k)^2}+\frac{1}{(2k+1)^2}\right)\tag{6}$$

We also notice that $a_k\to 0$ when $k\to\infty$.

It is obvious that series $S$ is convergent. Thus using the asymptotic property of the general term $b_n$ alone to decide the convergence is erroneous.