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. 

 A: Do not see Lebesgue dominated convergence in here. 
Rather, you could try to compute the distribution of $Z_n$ for each $n$ (hint: this is gaussian) and then show that $\mathrm P(|Z_n|\leqslant z)\to0$ for every fixed $z$ (hint: write each $Z_n$ as $a_nY_n+b_n$ with $Y_n$ standard gaussian and try to include each event $[|Z_n|\leqslant z]$ in an event $[|Y_n|\leqslant y_n]$ with $y_n\to0$). 
If one of these steps causes you some trouble and once you will have shown what you tried, please ask for further indications.
Edit After some exchanges in the comments, we know that $Z_n=\sqrt{n}\sigma Y_n+n\mu$ hence $|Z_n|\leqslant z$ if and only if $|Y_n+\sqrt{n}\mu/\sigma|\leqslant z/(\sqrt{n}\sigma)$. For the condition that $|Z_n|\leqslant z$ to hold, $Y_n$ must belong to the interval centered at $-\sqrt{n}\mu/\sigma$ of length $2z/(\sqrt{n}\sigma)$. Since the density of $Y_n$ is uniformly bounded by $1/\sqrt{2\pi}$, this shows that
$$
\mathrm P(|Z_n|\leqslant z)\leqslant2z/(\sqrt{2\pi n}\sigma)\to0.
$$
Alternatively, one could have noted from the start that the density of $Z_n$ is uniformly bounded by $1/(\sqrt{2\pi n}\sigma)$ and that one asks $Z_n$ to be in an interval of length $2z$.
In any case, if $Z_n\to \zeta$ in distribution and if $\zeta$ is not almost surely infinite, there exists $z$ such that $\mathrm P(|\zeta|\leqslant z)\geqslant\frac12$, say, hence $\mathrm P(|Z_n|\leqslant z)$ cannot converge to zero. This shows $Z_n$ cannot converge in distribution to a random variable $\zeta$ such that $\mathrm P(|\zeta|=\infty)\ne1$.
Note As this proof shows, my first indication (which was to include $[|Z_n|\leqslant z]$ in $[|Y_n|\leqslant y_n]$ with $y_n\to0$) was wrong since the suitable event involving $Y_n$ is NOT of the form $[|Y_n|\leqslant y_n]$ with $y_n\to0$ but $[|Y_n+t_n|\leqslant y_n]$ with $y_n\to0$ and $|t_n|\to\infty$. I leave everything as it was first written.
A: 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]
