Why do oscillating sequences diverge?

"Diverge" doesn't mean "grow big": it means "doesn't converge". In this case your sequence clusters around the two separate points $3$ and $3/2$, called limit points, rather than converging to a single point.

In the typical calculus setting, the behavior of a limit like $\lim_{n \to \infty} n^2$ would be more appropriately viewed as converging to $+\infty$ rather than as diverging, although for whatever reason this point of view isn't taught in introductory courses.

Because $$ \lim_{n\to\infty}a_{2n+1}=3 \text{ and } \lim_{n\to\infty}a_{2n}=\frac32 $$ and we have two subsequences convergent to different limits.

A rather more verbose way to explain it:

In a sequence which converges to some limit (let's call it $L$), if you go far enough along the sequence (to large enough $n$), every number in the sequence from that point on is "close enough" to $L$. But what counts as "close enough"? Well, the whole point of convergence is that it doesn't matter. No matter what threshold you choose for "close enough", there is some point at which all remaining numbers in the sequence are "close enough" to $L$.

The sequence you're looking at doesn't have that property. If you try $L = 3$, then it's not the case that all numbers past a certain point are close enough to $3$. I could choose "close enough" to be "within $0.1$" (for example) and then you are stuck with an infinite number of $\frac{3}{2}$s that are not close enough to $3$. Or if you try $L = \frac{3}{2}$, it's the opposite: you're stuck with an infinite number of $3$s that are not close enough to $\frac{3}{2}$. Hopefully you can tell that if you try any other number as $L$, either the $3$s or the $\frac{3}{2}$s or both will not be close enough to $L$.

So the sequence doesn't converge. That's precisely the definition of a divergent sequence: one that doesn't converge.

Do you know this:

$$\lim_{n\to\infty }x_n=A \iff \lim_{n\to\infty }x_{2n}=A= \lim_{n\to\infty }x_{2n+1}$$