It converges conditionally and not absolutely because $\displaystyle \sum_{n=2}^\infty \left|(-1)^n\cdot \dfrac{1}{\ln n}\right| = \displaystyle \sum_{n=2}^\infty \dfrac{1}{\ln n} \geq \displaystyle \sum_{n=2}^\infty \dfrac{1}{n} = \infty$, and you've done the first part that series is convergent by the alternating series test.
You can use a bunch of different tests, to get a variety of answers - each of them correct, but not necessarily equally informative.
In order from worst to best:
The alternating series test. This says that a series converges if (not only if!) the terms go to $0$, alternate sign, and have decreasing absolute value. The $n$th term in the series is $$a_n=(-1)^n\cdot ({4\over 9})^n\cdot{1\over 9},$$ so the alternating series test tells us that this converges.
The root test. While the alternating series test told us that the series converges, it didn't tell us whether it converged absolutely or merely conditionally. If we apply the root test, we will figure this out: we have $$\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert}=\lim_{n\rightarrow\infty}{4\over 9}\cdot {1\over\sqrt[n]{9}}=0,$$
so the series converges absolutely.
The ratio test. We have $\lim_{n\rightarrow\infty}{a_n\over a_{n+1}}=-{9\over 4}$, so - just like the root test - the ratio test says that the series converges absolutely. Like the root test, this provides more information than the alternating series test (which just said that the series converged).
But think about what we just did: the ratio of successive terms doesn't just approach $-{9\over 4}$, it always is ${9\over 4}$. That is, we're dealing with a geometric series whose ratio is $-{4\over 9}$ (which is strictly between $-1$ and $1$). This tells us not only that the series converges absolutely, but $\color{red}{\mbox{the exact value it converges to}}$ - namely, $${a_1\over 1-r}={-{4\over 81}\over {13\over 9}}=-{4\over 117}.$$ This is the right thing to do in this case. It gives the most information, and is as easy if not easier than anything else you can do here.
(Incidentally, the only reason I ranked the ratio test better than the root test is that it points out that the series is geomteric, if one has not already noticed that.)
Best Answer
You are wrong in your assumption that from the ratio of convergence you deduce that the series converge if and only if $-1<x\leqslant1$. The ratio test is inconclusive when $x=1$ and that's why you need the Leibniz test to determine whether the series converges if $x=1$.