Asymptotics of the maximum of quantized standard normals This is a problem from my measure theoretic probability class.  
Problem:
Given independent standard normals $Z_1,...Z_n$, let $X_i$ be the nearest integer to $Z_i$. Let $M_n$ be the maximum of $\{X_i,..., X_n\}$. Show that there exists an integer sequence $\{a_n\}$ and a sequence of probabilities $\{p_n\}$ such that $P(M_n = a_n) \sim  p_n$ and $P(M_n = a_n + 1) \sim 1 - p_n$. The symbol '$\sim$' means that the ratio of the left and right tends to 1 as $n$ goes to infinity.
Show that $p_n$ does not converge as $n$ goes to infinity.
Context:
The professor assigned this problem after we discussed how the CDF of the maximum of standard normals can asymptotically be written as a double exponential. He took $x = \sqrt{ 2 \log(n) - \log(\log(n)) + c }$ in order to show that the asymptotic distribution of the max of gaussians is $\exp(- \exp(-c/2) / (2 \sqrt{2 \pi}))$. He then went on to say that for discrete random variables "everything breaks."
My ideas:
I'm not really sure where to start, especially considering that the examples from class were all for continuous random variables. I think I understand the gist of the proposition: asymptotically the maximum of the $X_i$ will be in a window of two integers wp 1. The choice of $x$ for the continuous case is exotic and I think that the discrete case will require similar "magic."
Any hints or ideas? 
Thanks!
 A: One looks for an integer sequence $(a_n)$ such that, when $n\to\infty$, $P[M_n\leqslant a_n-1]\to0$ and $P[M_n\leqslant a_n+1]\to1$. 
For every $x$, $P[M_n\leqslant x]=\Phi(x+\frac12)^n$. When $x\to+\infty$, $\Phi(x)\to1$ hence 
$$
\log\Phi(x)\sim-(1-\Phi(x))\sim-1/(\sqrt{2\pi}x\mathrm e^{x^2/2}).
$$
Thus, one looks for some integer sequence $(a_n)$ such that $x\mathrm e^{x^2/2}\ll n$ for $x=a_n-\frac12$ and $x\mathrm e^{x^2/2}\gg n$ for $x=a_n+\frac32$. Equivalently, one asks that
$$
a_n\mathrm e^{a_n^2/2-a_n/2}\ll n\ll a_n\mathrm e^{a_n^2/2+3a_n/2},
$$
that is,
$$
\tfrac12a_n^2+\tfrac12a_n\pm a_n+\log a_n-\log n\to\pm\infty.
$$
Assume that 
$$
a_n=\lfloor\sqrt{2\log n}\rfloor,
$$
then $\sqrt{2\log n}-1\leqslant a_n\leqslant\sqrt{2\log n}$ hence
$$
\tfrac12a_n^2-\tfrac12a_n\leqslant \log n-\tfrac12\sqrt{2\log n}+O(1),
$$
and
$$
\tfrac12a_n^2+\tfrac32a_n\geqslant \log n+\tfrac12\sqrt{2\log n}+O(1),
$$
hence the sequence $(a_n)$ yields the result.
The fact that $a_n$ stays constant during longer and longer intervals, then jumps to $a_n+1$, and similar estimates to those above, probably entail that $\liminf\limits_{n\to\infty}p_n=0$ and $\limsup\limits_{n\to\infty} p_n=1$.
A: Since "everything breaks" for discrete random variables, do this in terms of the continuous random variables $Z_j$.  Start by relating $M_n$ to the maximum of the $Z_j$.
