This community wiki solution is intended to clear the question from the unanswered queue.
First off, failing the alternating series test doesn't imply that the series doesn't converge.
The example you gave is perfect: $\displaystyle a_n = \begin{cases}1/n^2 &\text{if $n$ is odd}, \\ 0 & \text{if $n$ is even}. \end{cases}$.
Notice that $\displaystyle \lim_{n \to \infty} a_n = 0$ but $a_n$ is not monotonically decreasing (it jumps up and down from $1/n^2$ to $0$).
However $\displaystyle \sum_{k = 1}^\infty a_n < \sum_{k = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} < \infty$ and hence converges.
Now, to disprove the statement in question, define $\displaystyle a_n = \begin{cases}1/n^2 &\text{if $n$ is odd}, \\ 1/n & \text{if $n$ is even}. \end{cases}$.
It is immediate that $\displaystyle \lim_{n \to \infty} a_n = 0$.
Notice that $$\displaystyle \sum_{n \text{ odd}} a_n = \sum_{n \text{ odd}} \frac{1}{n^2}< \sum_{k = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ and $$\displaystyle \sum_{n \text{ even}} a_n = \sum_{n \text{ even}} \frac1n = \frac12 \sum_{k = 1}^\infty \frac1n = \frac12 \cdot \infty = \infty.$$ Therefore $$\displaystyle \sum_{n = 1}^\infty (-1
)^n a_n = \sum_{n \text{ even}} a_n - \sum_{n \text{ odd}} a_n = \infty - \text{ finite positive number} = \infty.$$