Let $(B_{t})_{t\geq0}$ be a Brownian Motion. I would like to prove that $\max_{n\leq s\leq n+1}\left|\frac{B_{s}-B_{n+1}}{n}\right|=\frac{1}{n}\max_{n\leq s\leq n+1}\left|B_{s}-B_{n+1}\right|$ converges to $0$ in $L^{2}$ as $s\rightarrow\infty$ (and hence $n\rightarrow\infty$).
That is, I need to show that $\lim_{s\rightarrow\infty}\frac{1}{n^{2}}\mathbb{E}\left[\left(\max_{n\leq s\leq n+1}\left|B_{s}-B_{n+1}\right|\right)^{2}\right]=0$.
The only idea I have, is to somehow use the Reflection Principle, which gives us the distribution of $\max_{n\leq s\leq n+1}B_{s}$. I would be very thankful if anyone could help.
For every $n$, let $X_n=\max_{n\leq s\leq n+1}\left|B_{s}-B_{n+1}\right|$, then $(X_n)$ is identically distributed. Furthermore, $X_1$ is square integrable hence $$\lim\limits_{n\to\infty}\frac1{n^2}E(X_n^2)=\lim\limits_{n\to\infty}\frac1{n^2}E(X_1^2)=0.$$ Note that the quantity $\lim\limits_{\color{red}{s\to\infty}}\frac1{n^2}E(X_n^2)$ is undefined since $\frac1{n^2}E(X_n^2)$ does not depend on $s$.