Can you help me with the proof of the following theorem:
$X_n$ ~ $N(\mu_n, \sigma_n^2)$ are weak convergent to $X$ ~ $N(\mu, \sigma)$, if and only if $\lim_n \mu_n = \mu$ and $\lim_n \sigma_n = \sigma$, i.e.
$$X_n \rightarrow^D X \Leftrightarrow \lim_n \mu_n = \mu \text{ and } \lim_n \sigma_n = \sigma$$
Recall that a sequence $(X_n)_{n \in \mathbb{N}}$ of random variables converges to a random variable $X$ in distribution if and only if
$$\phi_{X_n}(\xi) := \mathbb{E}e^{i \xi X_n} \xrightarrow[]{n \to \infty} \phi_X(\xi) := \mathbb{E}e^{i \xi X} \tag{1}$$
for all $\xi \in \mathbb{R}^d$. Since $X_n \sim N(\mu_n,\sigma_n)$ and $X \sim N(\mu,\sigma^2)$, we have
$$\phi_{X_n}(\xi) = \exp \left( i \mu_n \xi - \frac{1}{2} \sigma_n^2 \xi^2 \right) \quad \text{and} \quad \phi_X(\xi) = \exp \left( i \mu \xi - \frac{1}{2} \sigma^2 \xi^2 \right). \tag{2}$$
If $\mu_n \to \mu$ and $\sigma_n \to \sigma$ it is obvious from $(2)$ that $(1)$ holds, and so $X_n$ converges in distribution to $X$. On the other hand, if $(1)$ holds, then in particular
$$|\phi_{X_n}(1)| \to |\phi_X(1)|,$$
and now it follows from $(2)$ that $\sigma_n^2 \to \sigma^2$. This, in turn, implies that
$$e^{i \mu_n \xi} \stackrel{(2)}{=} \exp \left( \frac{1}{2} \sigma_n^2 \xi^2 \right) \phi_n(\xi) \xrightarrow[]{n \to \infty} \exp \left( \frac{1}{2} \sigma^2 \xi^2 \right)\phi_X(\xi) \stackrel{(2)}{=} e^{i \mu \xi}$$
for all $\xi \in \mathbb{R}$. Hence, $\mu_n \to \mu$.