Question.
I am having difficulty working through the limit superior part of the following derivation of an asymptotic upper bound on the $(1 - \alpha)$ quantile on the probability of the most frequent number in roulette. I only have just come across its definition in real analysis (at the level of Abbott). I would appreciate if someone could help me clarify my confusion on:
Whether my attempt to reconstruct the derivation between $(13.35)$ and $(13.36)$ is correct (because it doesn't feel that way).
Why it is my attempt has led to a limit converging to something containing $n$.
Context.
The following is extracted from the The Doctrine of Chances: Probabilistic Aspects of Gambling, Ethier, S. (2010),
It follows from inclusion-exclusion-inequalities (Theorem 1.19 on p. 13) that$$[...] \leq \mathbb{P}\left(\max_{1 \leq i \leq r} X_i \geq l\right) \leq r\mathbb{P}(X_1 \geq l). \tag{13.35}$$Now $X_1$ is binomial$(n, 1/r)$, so we can apply DeMoivre's limit theorem (Theorem 1.5.13 on p.48) to conclude from the upper bound in $(13.35)$ that$$\limsup_{n \rightarrow \infty} \mathbb{P}\left(\max_{1 \leq i \leq r} X_i \geq l^+_n \right) \leq \alpha. \tag{13.36}$$if$$l^+_n := \frac{n}{r} + \Phi^{-1}\left(1 - \frac{\alpha}{r}\right)\sqrt{\frac{n}{r} \left(1 - \frac{1}{r}\right)}.$$
Where $(X_1, \dots, X_r) \sim \text{Multinomial}(n, 1/r, \dots, 1/r)$, so that $(X_1, \dots X_r)$ has multinomial distribution with sample size $n$ and $r$ cells, each with uniform probability $1/r$. This is the model for an unbiased generalised roulette wheel with $r \geq 36$ numbers, $n$ the number of spins, and $X_r$ the frequency of the number $r$ in $n$ spins.
Attempted derivation.
To proceed, I used the definition of the limit superior as the limit of a sequence of suprema. For some sequence $(x_n) = (x_1, x_2, \dots)$ that is bounded above, where all $x_i \in \mathbb{R}$, there will exist a "global" supremum $s$. Denote the sequence of suprema $(s_n) = \sup \{x_m \mid m \geq n\}$. Then,
$$\limsup_{n \rightarrow \infty} \ (x_n) = \lim_{n \rightarrow \infty} (s_n) = \lim_{n \rightarrow \infty} \left(\sup \{x_m \mid m \geq n\}\right).$$
Now because inequality $(13.35)$ applies for $l = 0, 1 \dots, n$, so assuming $l$ is now fixed, we can index the left and right hand side of $(13.35)$ with $n$,
$$\mathbb{P}_n\left(\max_{1 \leq i \leq r} X_i \geq l\right) \leq r\mathbb{P}_n(X_1 \geq l).$$
Consider the left hand side $\mathbb{P}_n\left(\max_{1 \leq i \leq r} X_i \geq l\right)$ as a sequence in $n$, with a view to showing that its supremum exists. We know the probability $\mathbb{P}\left(\max_{1 \leq i \leq r} X_i \geq l\right) \in [0, 1]$ for all $n$, and so the sequence $\mathbb{P}_n\left(\max_{1 \leq i \leq r} X_i \geq l\right)$ is also bounded above, and will have a supremum.
Similarly, for the right hand side, $\mathbb{P}(X_1 \geq l) \in [0, 1]$ for all $n$ and so $r\mathbb{P}_n(X_1 \geq l) \in [0, r]$, meaning that this sequence is also bounded above, and will have a supremum.
Now I consider the sequence $(s_n) = \sup\{r \mathbb{P}_m(X_1 \geq l) \mid m \geq n\}$, with a view to finding its limit. After doing some plotting, it appears that $(s_n) = (s_1, s_2, \dots) = (s_1, s_1, \dots)$, where
$$s_1 = \lim_{n \rightarrow \infty} r \mathbb{P}_n(X_1 \geq l)$$
With a view to evaluating this limit using DeMoivre Laplace, we have
$$\begin{aligned}r \mathbb{P}_n(X_1 \geq l) &= r(1 - \mathbb{P}_n(X_1 \leq l))\\&= r \left[1 - \mathbb{P}_n \left(Z_n \leq \frac{l - \frac{n}{r}}{\sqrt{\frac{n}{r}\left(1 - \frac{1}{r}\right)}}\right) \right] \\&= r - rF_{Z_n}\left(\frac{l - \frac{n}{r}}{\sqrt{\frac{n}{r}\left(1 - \frac{1}{r}\right)}}\right)\end{aligned}$$
Now using the fact that $\lim_{n \rightarrow \infty}F_{Z_n}(t) \rightarrow F_Z(t)$ for $Z \sim N(0,1)$, I have that
$$s_1 = \lim_{n \rightarrow \infty} r \mathbb{P}_n(X_1 \geq l) = r - r \Phi\left(\frac{l - \frac{n}{r}}{\sqrt{\frac{n}{r}\left(1 - \frac{1}{r}\right)}}\right).$$
We now have
$$\begin{aligned}\limsup_{n \rightarrow \infty} \mathbb{P}\left(\max_{1 \leq i \leq r} X_i \geq l \right) &\leq \lim_{n \rightarrow \infty} \left(s_1, s_1, s_1, \dots \right) \\&= r - r \Phi\left(\frac{l - \frac{n}{r}}{\sqrt{\frac{n}{r}\left(1 - \frac{1}{r}\right)}}\right) \\\end{aligned}$$
Whilst I can see that from here we just invert the probability inequality, what I don't understand is how it is we can have the result of the limit on the R.H.S contain $n$?
I have a feeling I have gone completely off the rails here, and it may be to do with the fact that I have misunderstood asymptotic approximations like DeMoivre Laplace as us having faith that $n$ is large enough to approximate the relevant Binomial probability, meaning that we don't have a resulting limit that "contains n".