Show $\min_{0\le k \le n}P(|S_n-S_k|\le n\epsilon)\to 0$ If $S_n/n\to 0$ in probability, How to show that $$\min_{0\le k \le n}P(|S_n-S_k|\le n\epsilon)\to 1$$ as $n\to \infty$?
By definition of converge in probability,  $P(|S_n|> n\epsilon)\to 0$, so $P(|S_n|\le n\epsilon)\to 1$ as $n\to\infty$. Then how to proceed since $|S_n-S_k|\le |S_n|+|S_k|$?
Thanks!
 A: By considering the complement of the event $\left\{|S_n-S_k|\le n\epsilon\right\}$, the problem is equivalent to the following one: show that for any positive $\epsilon$, 
$$\lim_{n\to +\infty} \max_{1\leqslant k\leqslant n} \mathbb P\left(\left\{\left|S_n-S_k\right|\gt n\epsilon \right\}\right)=0.$$
Note that $$\mathbb P\left(\left\{\left|S_n-S_k\right|\gt n\epsilon \right\}\right)\leqslant \mathbb P\left(\left\{\left|S_n\right|\gt n\epsilon/2 \right\}\right)+\mathbb P\left(\left\{\left|S_k\right|\gt n\epsilon /2\right\}\right),$$ 
hence it suffices to prove that 
$$\lim_{n\to +\infty} \max_{1\leqslant k\leqslant n} \mathbb P\left(\left\{\left|S_k\right|\gt n\epsilon \right\}\right)=0.$$
To this aim, we notice that for any fixed $R$ and any $n\geqslant R$, 
$$\max_{1\leqslant k\leqslant n} \mathbb P\left(\left\{\left|S_k\right|\gt n\epsilon \right\}\right)\leqslant\max_{R\leqslant k\leqslant n} \mathbb P\left(\left\{\left|S_k\right|\gt k\epsilon \right\}\right)+
\max_{1\leqslant k\leqslant R} \mathbb P\left(\left\{\left|S_k\right|\gt n\epsilon \right\}\right).$$
Since the sequence $\left( \mathbb P\left(\left\{\left|S_k\right|\gt k\epsilon \right\}\right)\right)_{k\geqslant 1}$ goes to $0$ as $n$ goes to infinity, we can choose $R$ such that the contribution of $\max_{R\leqslant k\leqslant n} \mathbb P\left(\left\{\left|S_k\right|\gt k\epsilon \right\}\right)$ is negligible for any $n$. 
