Maximum of a sequence is $o(\sqrt{n})$ Quick context: This is a transposition of exercise 6.3 of Remco Van der Hofstad (2016) and it is relevant to some problem i encoutered in my research.
For each $n \in \mathbb{N}$, define a series of positive numbers $w_i^n$ with $1\leq i\leq n$. Also denote an uniformly chosen numbers by $w_U = W_n$ where $U$ is picked uniformly in $\{1,...,n\}$ (meaning $\mathbb{P}(W_n = w_i^n) = 1/n$ for any $i\leq n$). We also suppose that :
a)$W_n$ converges in distribution to a positive variable in the sense that  $\lim_{n\rightarrow\infty}\mathbb{P}(W_n \leq t) = \mathbb{P}(W \leq t)$
b) $\lim_{n\rightarrow\infty} \mathbb{E}W_n = \mathbb{E}W$
c) $\lim_{n\rightarrow\infty} \mathbb{E}W_n^2 = \mathbb{E}W^2$
The question is to show that A. if a) and b) hold then $\max_{i\leq n} w_i^n = o(n)$ and that B. if a), b) and c) then $\max_{i\leq n} w_i^n = o(\sqrt{n})$.
As for my approach, I have noticed A. and B. are proven the same way. To prove A. I wanted to use the Borel Cantelli Lemma to show that $w_i^n/n > \epsilon$ only a finite amount of times. We have :
\begin{equation}
\sum_{ i\leq n } 1_{ w_i^n/n > \epsilon } = \sum_{ i\leq n} 1_{w_i^n > n\epsilon}
\end{equation}
which can be rewritten as
\begin{equation}
\sum_{ i\leq n } 1_{ w_i^n/n > \epsilon } = \sum_{ d = \lceil n\epsilon \rceil } d\mathbb{P}( W^n = d) 
\end{equation}
The rightmost part can be majored by something close to the expectation of $W$ regardless of $n$. My intuition tells me that taking $n$ to $\infty$ feel like it should be enough to finish the proof but i feel like i did not use all the hypotheses... I'm open to any ideas about the proof.
PS: I asked an adjacent question yesterday which might be relevant.
 A: Fix a positive number $L>0$. You can write
$$
\max_{i\leq n} w_i^n \leq \max\left(L,\sum_{i=1}^nw_i^n \mathbf{1}(w_i^n >L)\right).
$$
The above inequality is true because either 1. $\max_{i\leq n}w_i^n \leq L$ or 2. $\max_{i\leq n}w_i^n$ is bounded above by $\sum_{i=1}^nw_i^n \mathbf{1}(w_i^n >L)$ as $w_i^n\geq 0$. Now use the fact that if $a,b\geq 0$ then $\max(a,b)\leq a+b$ to get
$$
\frac{1}{n}\max_{i\leq n} w_i^n \leq \frac{L}{n} + \frac{1}{n}\sum_{i=1}^nw_i^n \mathbf{1}(w_i^n >L)  = \frac{L}{n} + \mathbb{E}\left(W_n \mathbf{1}(W_n >L)\right).
$$
Now $\mathbb{E}\left(W_n \mathbf{1}(W_n >L)\right)= \mathbb{E}\left(W_n\right) - \mathbb{E}\left(W_n \mathbf{1}(W_n \leq L)\right)$. You are given that $\mathbb{E}(W_n) \to \mathbb{E}(W)$, and since $W_n \mathbf{1}(W_n \leq L) \leq L$, and $W_n$ converges to $W$ in distribution, you get $\mathbb{E}\left(W_n \mathbf{1}(W_n \leq L)\right) \to \mathbb{E}\left(W \mathbf{1}(W \leq L)\right)$, as $n\to \infty$. Therefore $\mathbb{E}\left(W_n \mathbf{1}(W_n >L)\right) \to \mathbb{E}\left(W \mathbf{1}(W >L)\right)$. Thus
$$
\limsup_{n\to \infty} \frac{1}{n}\max_{i\leq n} w_i^n\leq \mathbb{E}\left(W \mathbf{1}(W >L)\right).
$$
Now noting that $W \mathbf{1}(W >L) \leq W$, you can use dominated convergence theorem and let $L \to \infty$ and get $\limsup_{n\to \infty} \frac{1}{n}\max_{i\leq n} w_i^n = 0$, which gives $\max_{i\leq n} w_i^n = o(n)$.
If $\lim_{n\rightarrow\infty} \mathbb{E}W_n^2 = \mathbb{E}W^2$, then note that (using similar reasoning as before)
$$
(\max_{i\leq n} w_i^n )^2\leq \max\left(L^2,\sum_{i=1}^n(w_i^n)^2 \mathbf{1}(w_i^n >L)\right).
$$
Again, by an almost identical argument as before
$$
\frac{1}{n}(\max_{i\leq n} w_i^n)^2 \leq  \frac{L^2}{n} + \mathbb{E}\left(W_n^2 \mathbf{1}(W_n >L)\right) \to \mathbb{E}\left(W^2 \mathbf{1}(W >L)\right).
$$
Now again taking $L\to \infty$, get $(\max_{i\leq n} w_i^n)^2 =o(n)$ and therefore $\max_{i\leq n} w_i^n =o(\sqrt{n})$.
