Series convergence question

**The question:**

Determine if the following alternating series converges. Is it absolutely convergent?

$\displaystyle \sum\limits_{k = 3}^\infty \frac{(-1)^k}{ln(k)}$

**My attempt:**

Using the alternating series test, we need:

a) $\displaystyle a_k \ge 0$

b) $\displaystyle a_{k} \ge a_{k + 1}$

c)$\displaystyle \lim\limits_{k \to \infty} a_k = 0$

a) $\displaystyle \frac{1}{ln(k)}$ > 0 for all k > 1

b) $\displaystyle \frac{1}{ln(k)} \ge \frac{1}{ln(k+1)}$ for all k > 1

c) Limit is obviously 0 as $\displaystyle k \to \infty$

Thus the series converges.

Now we consider $\displaystyle \sum\limits_{k = 3}^\infty |\frac{1}{ln(k)}|$

Using ratio test:

$\displaystyle r = \frac{a_{k + 1}}{a_k} = |\frac{1}{ln(k + 1)}}| |\frac{ln(k)}{1}}|$

= $\displaystyle \frac{ln(k)}{ln(k+1)} < 1$

Therefore converges. So this series should be absolutely convergent. However the answer is actually 'conditionally convergent'.

Where have I gone wrong? Thanks.