$$\frac{1}{\sqrt 1} + \frac{2}{\sqrt 2} - \frac{3}{\sqrt 3} + \frac{1}{\sqrt 4} + \frac{2}{\sqrt 5} - \frac{3}{\sqrt 6} + \frac{1}{\sqrt 7} + \frac{2}{\sqrt 8} - \frac{3}{\sqrt 9} + \dots $$
By what my class has been going over, I assume I can either reorder this sequence, which then I am not sure what to do with it.
I would caution against rearranging series to determine convergence. It is a theorem due to Riemann that any conditionally converging series can be rearranged to converge to any real number or even to diverge. One can also show that there exist diverging series which can be rearranged to converge (Can a divergent alternating series by rearrangement of terms be made to converge to a value?).
The series you've written down, if it converges, is fact conditionally convergent. To see this, notice that $$\frac{1}{\sqrt{1}} + \frac{2}{\sqrt{2}} + \frac{3}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \frac{2}{\sqrt{5}} + \frac{3}{\sqrt{6}} + \cdots$$ is strictly larger than the series $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{6}} + \cdots$$ which is known to diverge. Therefore, by the comparison test, the series formed by taking the absolute value of each term in the original series diverges. That is, the original series, if it converges, is conditionally convergent.
This means that we can't analyze the convergence of the series by rearranging its terms. Instead, I recommend applying Dirichlet's test for convergence. It says that, given sequences of real numbers $(a_n)$ and $(b_n)$, if:
1) The partial sums of $(a_n)$ are bounded
2) $(b_n)$ is decreasing
3) $\lim_{n\to \infty} b_n = 0$
then $\sum a_n b_n$ converges. Take $(a_n) = (1,2,-3,1,2,-3, \ldots)$ and $b_n = \frac{1}{\sqrt{n}}$ and check that the conditions of Dirichlet's theorem are met. Note that this test doesn't tell you what the sum converges too, only that it does in fact converge.
We can also prove convergence by means of the Leibniz criterion. Your sequence $(a_n)$ has the form $a_{3k+j} = (-1)^{\delta_{3,j}}\tfrac{j}{\sqrt{3k+j}}$, $k\ge 0$, $j=1,2,3$. Set $A_n := \sum_{k=1}^na_k$.
Now, define $b_{2k+1} = a_{3k+1}+a_{3k+2}$ and $b_{2k+2} = |a_{3k+3}|$, $k\ge 0$. Then $b_{2k+1}\ge b_{2k+2}$ and $b_{2k+2}\ge b_{2(k+1)+1}$. So, $(b_k)$ decreases and it obviously tends to zero. Now, put $B_n := \sum_{k=1}^n(-1)^{k+1}b_k$. Then $(B_n)$ converges by Leibniz to some value $S$. Moreover, we have $A_{3n+2} = B_{2n+1}\to S$ and $A_{3n+3} = B_{2n+2}\to S$. But since $A_{3n} < A_{3n+1} < A_{3n+2}$, it follows that also $A_{3n+1}\to S$ as $n\to\infty$. Hence, $A_n\to S$ as $n\to\infty$.
