Limiting distribution of sum of normals How would I go about solving this problem below? I am not exactly sure where to start. I know that I need to make use of the Lebesgue Dominated Convergence theorem as well. Thanks for the help.

Let $X_1, X_2, \ldots, X_n$ be a random sample of size $n$ from a distribution that is $N(\mu, \sigma^2)$ where $\sigma^2 > 0$. Show that $Z_n = \sum\limits_{i=1}^n X_i$ does not have a limiting distribution. 


Another way to show $P(Z_n\le z)=0$ for all $z$ as Did suggests:
Note that $Z_n\sim N(n\mu, n\sigma^2)$, so for some fixed real number $z$, $P(Z_n\le z)=\frac{1}{2}\biggr[1+erf\biggr(\frac{z-n\mu}{\sqrt{2n\sigma^2}}\biggr)\biggr]\to0$ as $n\to\infty$,
since $\frac{z-n\mu}{\sqrt{2n\sigma^2}}\to-\infty$ and $\lim_{x\to-\infty}erf(x)=-1$.
Hence the limit of the CDFs is equal to 0 everywhere, and cannot limit to a CDF.
[I use the definition of erf from the Wikipedia page: https://en.wikipedia.org/wiki/Normal_distribution]
