Suppose $S_n$ is a simple random walk; formally, $S_n=\sum_{i=1}^n X_i$ for $X_i\sim\mathcal{U}(-1,1)$, i.i.d.. Denote by $M_n$ the maximum of the random walk on $n$ steps; formally, $M_n=\max_{0\le k\le n}S_k$. Finally, let $T_n$ be that maximum's hitting time; formally, $T_n=\min\{k,\ 0\le k\le n:\ S_k=M_n\}$.
We may have dual notations for the standard Brownian motion $B(t)$ on the interval $0\le t\le 1$: $M$ is the maximum of the motion (formally, $M=\max_{0\le t\le 1}B(t)$), and $T$ is the hitting time of $M$ (formally, $T=\min\{t,\ 0\le t\le 1:\ B(t)=M\}$).
I know that at least in some sense, the random walk $S_n$ converges to the Brownian motion $B(t)$. Here's a way to formalize this: if $S(t)$ is the linear interpolation of $S_n$; that is $$S(t)=S_{\lfloor t\rfloor}+(t-\lfloor t\rfloor)(S_{\lfloor t\rfloor + 1} - S_{\lfloor t\rfloor}),$$ and $S_n^*(t) = S(nt)/\sqrt{n}$ then $S_n^*$ converges in distribution to $B(t)$ (on the interval $[0,1]$).
My question is as follows: does it follow from the above convergence, or true for any other reason, that $T_n/n$ converges in distribution to $T$?