Difficult demonstration - How to show that $H_n$ is normal distributed $N(\xi,\sigma^2)$ starting from its moments $ξ$ and $σ$?

I was thinking that if the function $$H_n$$ of cumulative distribution converges to a distribution $$H$$, then $$\epsilon_n$$ should converge to $$\epsilon$$ what could be expressed as follows:

If $$H_n$$ is the normal distribution with mean $$\epsilon_n$$ and variance $$\sigma_n^2$$, then $$H_n$$ tends to the normal distribution H with mean $$\epsilon$$ and variance $$\sigma^2>0$$ if and only if

$$\epsilon_n \to \epsilon$$ and $$\sigma_n^2 \to \sigma^2$$.

How could this defense be demonstrated?

A sequence of random variables $$H_n$$ defined on $$\mathbb{R}$$ converge in distribution to $$H$$ (denoted $$H_n \Rightarrow H$$) if and only if the sequence of cumulative distribution functions (c.d.f.s) $$F_n(x)$$ converge pointwise to the c.d.f. $$F$$ of $$H$$ at all points of continuity of $$F$$.
If $$\epsilon_n \rightarrow \epsilon$$ and $$\sigma_n \rightarrow \sigma$$, then $$\lim_{n\rightarrow\infty}F_n(x) = \lim_{n\rightarrow\infty} \int_{-\infty}^x \frac{e^{-(y-\epsilon_n)^2 / 2\sigma_n^2}}{\sqrt{2 \pi\sigma_n^2}} dy = \int_{-\infty}^x \frac{e^{-(y-\epsilon)^2}}{\sqrt{2 \pi \sigma^2}} dy = F(x),$$ where the interchange of limit and integral is permitted because the integrands converge uniformly. Thus $$F_n(x) \rightarrow F(x)$$ for all $$x$$, so $$H_n \Rightarrow H$$.
Conversely, if $$H_n \Rightarrow H$$, since $$F$$ is continuous everywhere for a Gaussian we have $$F_n(x) \rightarrow F(x)$$ for all $$x$$. Moreover, if $$F_{\epsilon, \sigma}$$ is the c.d.f. of a Gaussian with mean $$\epsilon$$ and variance $$\sigma^2$$, then $$F_{\epsilon_1,\sigma_1}(x) = F_{\epsilon_2,\sigma_2}(x)$$ for all $$x$$ if and only if $$\epsilon_1 = \epsilon_2$$ and $$\sigma_1 = \sigma_2$$. Since limits in distribution are unique, it must be that $$\lim_{n\rightarrow\infty} \epsilon_n$$ and $$\lim_{n\rightarrow\infty} \sigma_n = \sigma$$.
Convergence in distribution is equivalent to convergence of characteristic functions. So we have to show that $$e^{i\epsilon_n t} e^{-t^{2}\sigma_n^{2}/2} \to e^{i\epsilon t} e^{-t^{2}\sigma^{2}/2}$$ for all $$t$$ iff $$\epsilon_n \to \epsilon$$ and $$\sigma_n \to \sigma$$ which is quite easy: if the condition holds take absolute values to conclude that $$\sigma_n \to \sigma$$ and then show that $$\epsilon_n \to \epsilon$$.