An ancillary result from convergence in probability I was reading a paper concerning probability theory. 
We have that $X_i$, $i = 1,2,...$ i.i.d random variables, and $S_n = \sum_{i=1}^nX_i$ is defined as partial sum as usual. If 
$$\frac{S_n}{n} \to 0 \quad \text{  in probability}$$
then the author says that we can easily get 
$$\lim_{n\to \infty} \min_{1\leq k \leq n}\mathbb{P}\left(\frac{|S_n - S_k|}{n} < \epsilon\right) = 1 $$
for any $\epsilon > 0$. I tried to prove this but fail to give a rigorous proof. Really appreciate if any one could give some hints. Or if you don't agree with this, could you give a counter example to show that it does not hold. 
 A: Since
$$\left\{ \left| \frac{S_n-S_k}{n} \right|< \epsilon \right\} \supseteq \left\{ \left| \frac{S_n}{n} \right|< \frac{\epsilon}{2} \right\} \cap \left\{ \left| \frac{S_k}{n} \right| < \frac{\epsilon}{2} \right\}$$
we have
$$\mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) \geq \mathbb{P} \left( \left| \frac{S_n}{n} \right| < \frac{\epsilon}{2} \right) - \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right)$$
which implies
$$\begin{align*} \min_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) &\geq \mathbb{P} \left( \left| \frac{S_n}{n} \right| < \frac{\epsilon}{2} \right) - \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \\ &\geq 1- 2 \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right)  \end{align*}$$
Therefore, the assertion follows if we can show
$$\lim_{n \to \infty} \max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) = 0. \tag{1}$$
Fix $\delta>0$. Since $S_n/n \to 0$ in probability there exists $N_0 \in \mathbb{N}$ such that
$$\mathbb{P} \left( \left| \frac{S_n}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta \quad \text{for all $n \geq N_0$.} \tag{2}$$
Choosing $N_1 \geq N_0$ sufficiently large we can achive
$$\mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta \quad \text{for all $k \in \{1,\ldots,N_0\}$ and $n \geq N_1$}. \tag{3} $$
On the other hand,
$$\mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \mathbb{P} \left( \left| \frac{S_k}{k} \right| \geq \frac{\epsilon}{2} \right) \stackrel{(2)}{\leq} \delta \tag{4}$$
for all $k \in \{N_0+1,\ldots,n\}$ and $n \geq N_1 \geq N_0$. Combining (3) and (4) shows
$$\max_{1 \leq k \leq n} \mathbb{P} \left( \left| \frac{S_k}{n} \right| \geq \frac{\epsilon}{2} \right) \leq \delta$$
for all $n \geq N_1$. Since $\delta>0$ is arbitrary, this proves (1).
