Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables with the property that $N(m_n, \sigma_n^2)$. Prove that if $X_n$ converges in distribution to $X$ then $X \sim N(m,\sigma^2)$ where $m = \lim_{n \to \infty} m_n$ and $\sigma^2 = \lim_{n \to \infty} \sigma_n^2$.
We know by definition that the cumulative distribution function of the corresponding random variables converge at every continuity point. Now, using a result of Levy's theorem that convergence in distribution is equivalent with convergence of the corresponding characteristic functions, I want to show that the above statement holds. But for me it is unclear how the convergence of the characteristic functions results in $X \sim N(m,\sigma^2)$.