Let $(\mu_n)_{n \geq 1}$ be a sequence of real numbers and $(\sigma_n)_{n \geq 1}$ be a sequence of positive numbers, and let $\mu \in \mathbb{R}$ and $\sigma>0$. For $n \geq 1$, let $X_n\sim N(\mu_n, \sigma_n^2)$. Assume $\mu_n \rightarrow \mu$ and $\sigma_n \rightarrow \sigma$.
I need to prove that the $$X_n \rightarrow N(\mu, \sigma^2)$$ in distribution
I wanted to show that the limit hold in probability, however, I am not sure how to proceed
$X_n=_D\mu_n + \sigma_n Z$, $Z$ standard normal. Since $Z$ converges to $Z$ in distribution, Slutsky's theorem implies $X_n$ converges to $\mu+\sigma Z \sim N(\mu,\sigma^2)$ in distribution. You could also prove this with characteristic functions. Your statement about convergence in probability is true but more than you need if you're willing to use Slustky (or ch. function).