# Find convergence in probability for variance that changes

I'm preparing for my exam and was trying out this question:

Let $$X_1, X_2,\dots,X_n$$ be independent random variables with $$E(X_i) =\mu, V(X_i) = \begin{cases} 2 & \text{ if } i \text{ is odd} \\ 3 & \text {if } i \text{ is even}\end{cases}$$. Show that $$\bar{X} \stackrel{\Bbb{P}}\to \mu$$ as $$n \to \infty$$.

Here is my working: $$\Bbb{P}(|\bar{x}-\mu| \geq \epsilon) \leq \frac{\operatorname{\rm Var}(\bar{x})}{\epsilon^2}=\frac{2}{n\epsilon^2} \text{ or } \frac{3}{n\epsilon^2} \to 0 \text{ as } n \to \infty,$$ so as $$n \to \infty,\bar{X} \stackrel{\Bbb{P}}\to \mu$$.

I have a feeling that it is not correct. Anyone can help point out where I did wrong?

$$\text{Var}(\bar{X}_n) = \frac{1}{n^2} \sum_{i=1}^n \text{Var}(X_i) \le \frac{3}{n}.$$