$N(0,\sigma^2_n)$ and $\sigma^2_n\to\sigma^2$ imply $N(0,\sigma^2_n)\overset{d}{\to}N(0,\sigma^2)$? The following result seems to be natural to me:
$$N(0,\sigma^2_n) \text{ and } \sigma^2_n\to\sigma^2 \implies N(0,\sigma^2_n)\overset{d}{\to}N(0,\sigma^2)$$
as $n\to\infty$, where $\overset{d}{\to}$ denotes convergence in distribution. But I can't find the exact argument to show this.
Can you point me out?
My attempt
The moment generating function uniquely determines the distribution function. We can then stablish
$$\lim M_{X_n}(t)=M_X(t), \forall t\in (-a,a) \implies X_n\overset{d}{\to}X,$$
for some $a>0$.
As $\lim e^{\sigma^2_nt^2/2}=e^{\sigma^2t^2/2}$ , we have the result for the gaussian case.
I printed below, Curtiss's theorem where he stated the above result:

For the Gaussian case, the moment generating function exists for any real $t$ and $\lim_{n\to\infty} M_{X_n}(t)=M_X(t)$ holds for any real $t$, given that $\sigma^2_n\to\sigma^2$.
 A: Perhaps the fastest way is to use CF, that is $\varphi_n(t) = \exp (-\frac{t^2\sigma_n^2}{2}) \to \exp(-\frac{t^2\sigma^2}{2}) = \varphi(t)$, where $\varphi_n,\varphi$ are respectivelly CF of $\mathcal N(0,\sigma_n^2)$ and $\mathcal N(0,\sigma^2)$ distributions. Now use Levy Cramer continuity theorem and you have your result.
But your reasoning is correct, too. Note that convergence in distribution is equivalent with convergence of CDF at continuity point of limiting distribution. It is even enough that there exists $\delta > 0$ such that if $M_n(t) \to M(t)$ for $t \in (-\delta,\delta)$, because then we get $\lim F_n(t) = F(t)$ for every continuity point t of $F$, so if $M_n(t) \to M(t)$ for every $t$ in small interval with $0$ isnide, then corresponding distributions converge in the weak sense.
If $\sigma > 0$ we can apply sheffe thereom. Since densities converge a.s (note that for $n>N$ we would have $\sigma_n > 0$, so we have densities) then corresponding distributions converge, too (even in the stronger sense of total variation). 
Another way is to try to tackle it by definition, but it would be a bit problematic because of cases $\sigma = 0$ and $\sigma > 0$
