Proving estimator consistency I have the following estimator: $\hat{\sigma}^2_N = \frac{1}{h^2}\sum\limits_{i=1}^{N}x^2_i$, where $x_i \sim i.i.d. \; \mathcal{N}(\mu\frac{h}{N}, \sigma^2\frac{h}{N})$.
We can show that $E[\hat{\sigma}^2_N] = \sigma^2 + \mu^2 \frac{h}{N}$ and that $var[\hat{\sigma}^2_N] = 2\sigma^4 \frac{1}{N} + 4\mu^2 \sigma^2 \frac{h}{N^2}$.
I know that it is possible to prove that $\hat{\sigma}^2_N$ is a consistent estimator of $\sigma$, but I don't know how to proceed. 
My attempt is: since $E[\hat{\sigma}^2_N] \to \sigma^2$ as $N \to \infty$, $\hat{\sigma}^2_N$ is asymptotically unbiased. Moreover, we have that $var[\hat{\sigma}^2_N] \to 0$ as $N \to \infty$, hence the estimator is consistent.
I know it is not very rigorous, but is this reasoning valid? If not, how is it possible to prove that the estimator is consistent?

The estimator $\hat{\sigma}^2_N$ is consistent if it converges in probability to $\sigma^2$. To prove consistency it is sufficient to show that $E[(\hat{\sigma}^2_N - \sigma^2)^2]$ goes to $0$ as $N \to \infty$.
In order to show this, we exploit the fact that $E[(\hat{\sigma}^2_N - \sigma^2)^2] = E[(\hat{\sigma}^2_N)^2] -2\sigma^2E[\hat{\sigma}^2_N] + \sigma^4 = 2\sigma^4 \frac{1}{N} + 4\mu^2 \sigma^2 \frac{h}{N} + \mu^4 \frac{h^2}{N^2}$, which is equivalent to the sum of the variance and the bias. It is immediate to see that this quantity goes to $0$ as $N \to \infty$. Therefore we proved that the estimator is consistent.
