As the title implies, I need to check whether $\bar{X}^2$ is a consistent estimator of $\mu^2$ with $X\sim\mathcal{N}(\mu, \sigma^2)$. So far I've calculated the bias as follows: $$ E[\bar{X}^2]-\mu^2=V[\bar{X}]+E^2[\bar{X}] -\mu^2 = \frac{\sigma^2}{n} +\mu^2 - \mu^2 = \frac{\sigma^2}{n} $$ and the only thing left to do is to check whether the variance also tends to $0$ or not. But I'm having a hard time figuring it out, so far I've came up with this: $$ V[\bar{X}^2]=V[(\sum_{i=1}^{n}{\frac{X_i}{n}})^2]=\frac{1}{n^4}V[(\sum_{i=1}^{n}{X_i})^2] \\ \frac{1}{n^4}V[(\sum_{i=1}^{n}{X_i})^2]=\frac{1}{n^4}V[nX^2+(n^2-n)X_iX_j] \text{ where $x_i$ and $x_j$ are iid.} \\ \frac{1}{n^4}V[nX^2+(n^2-n)X_iX_j]=\frac{1}{n^4}(n^2V[X^2]+(n^2-n)^2V[X_iX_j]) $$ Now, that last expression approaches $V[X_iX_j]$ as $n$ tends to infinity, and that is equal to: $\sigma^4 + 2(\sigma\mu)^2$ (I skipped the equations to get there because they're tedious and I'm sure there's no error there).
But this means that the mean squared error tends to $\sigma^4 + 2(\sigma\mu)^2$ which is not necessarily equal to $0$, and that would mean that the estimator is inconsistent. However I find that conclusion to be really counter intuitive and I'm pretty sure I must have made a mistake somewhere. Could someone point out where I went wrong or tell me if my proof is correct?
$$V(\overline{X}^2) = \mathbb E[\overline{X}^4] - \mathbb E[\overline{X}^2]^2 $$ Since $X \sim \mathcal N(\mu, \sigma^2)$, $\overline{X} \sim \mathcal N(\mu, \sigma^2/n)$.
You should get $$V(\overline{X}^2) = 4\,{\frac {{\sigma}^{2}{\mu}^{2}}{n}}+2\,{\frac {{\sigma}^{4}}{{n}^{2} }} $$