Does the series $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$ converge?

Does the following variant of the harmonic series converge? If it diverges (which I think it does), can I know if it diverges to $\infty$ or has no limit?

Note that the series is not alternating in the classical sense of the word. $$1 + \frac{1}{2} -\frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$

The generic term of the series would have to be something like,

$$a_n = \left\{\begin{array}{ll} -\frac{1}{n}, & \text{if } 3 \mid n \\ \;\, \,\, \frac{1}{n}, & \text{otherwise}\end{array}\right.$$ I'm not sure if it's very helpful. The terms divisible by 3 are negative, and the others are positive.

Is there a way to decide and prove whether an alternating series of this sort (e.g. with a period other than 2) converges? Or one where terms are positive or negative according to some other rule?

Almost all convergence tests I've come across are generally limited to either simple alternating series or where all the terms are positive.

No, the series diverges. Note that the partial sum of length $3N$ can be written as

$$\sum_{n=0}^{N-1} \left(\frac{1}{3n+1} + \frac{1}{3n+2} - \frac{1}{3(n+1)}\right).$$

Combining the terms, we get

$$\sum_{n=0}^{N-1} \frac{9 n^2+18 n+7}{3 (n+1) (3n+1) (3 n+2)}$$

and it's easy to see that this is a divergent series by limit comparison with $\sum 1/n$. Now, if the original series converged, then this sub-sequence of the sequence of partial sums would also have to converge.

Note that the partial sums $S_n \geq 1 + \frac{1}{4} + \frac{1}{7} + \ldots = \sum_{k=0}^{[n/3]}\frac{1}{3k+1}$, bounded from below by some partial sum of the harmonic series (divided by $3$), which diverge to infinity.