Resnick - Probability Path - Exercise 6.16 (c) I'm trying to solve the following exercise from Resnick's books:

For any sequence of random variables {$X_n$} set $S_n = \sum_{i=1}^{n}X_i$
(c) Show $X_n \xrightarrow{P} 0$ does NOT imply $S_n/n \xrightarrow{P} 0$.
Hint: Try $X_n = 2^n$ with probability $n^{-1}$ and $=0$ with probability  $1-n^{-1}.$

I could show the first part, that $X_n \xrightarrow{P} 0$, by taking:
$\lim_{n\to\infty}P(X_n = 0) = \lim_{n\to\infty}(1-\frac{1}{n}) = 1$
$\Rightarrow \lim_{n\to\infty}P(|X_n - 0|\leq \epsilon) =1 $
$\Rightarrow \lim_{n\to\infty}P(|X_n - 0|> \epsilon) =0 $, by taking complements.
How can I show the second part, that $S_n/n \xrightarrow{P} 0$ does not apply?
 A: I would argue as follows. We show that:
$$\limsup_{n \to \infty} \mathbb{P}(|S_n|  \ge n) \ge 1 - \frac{1}{e}.$$ Indeed fix $n = 2^l$ for some integer $l$ (in the end we will take $l \to \infty$). Then 
$$
\mathbb{P}(|S_n| \ge n) \ge \mathbb{P}( \exists m \ge l, \text{ such that } X_m \neq 0) = 1 -\Big( 1 -\frac{1}{n} \Big)^{K(n)}
$$
Indeed the first inequality follows since for such $m$ we would get $S_n/n \ge X_m/n \ge 1.$ The latter equality follows on the other side by counting the probability of at least one success in a sequence of Bernoulli trials. $K(n)$ indicates the number of trials, namely the number of admissible $m$, that is:
$$K(n) = 
\#\{m: \ \ l \le m  \le  2^l \} = 2^l {-} l{+}1.$$
Hence $K(n) \simeq n.$ Passing to the limit (over $l$ or $n$) on the right hand side gives then the solution.
A: For any given $\epsilon>0$, arbitrarily close to $0$, we would have that:
$P(|\frac{S_n}{n}|>\epsilon)=P([X_{n_{0}}=0,X_{n_{0}+1}=0,\dots,X_n=0]^c)= 1-(1-\frac{1}{n})^{n-n_0}$
So that $n_0$ is the minimum value such that $X_i\geq n$, to guarantee that for any $\epsilon$ is true that $\frac{S_n}{n}>\epsilon$.
Now, taking limits both sides
$\lim_{n\to\infty}P(|\frac{S_n}{n}|>\epsilon)=\lim_{n\to\infty}[1-(1-\frac{1}{n})^{n-n_0}] =1- \lim_{n\to\infty}(1-\frac{1}{n})^{n-n_0} = 1-\frac{1}{e} \neq 0$
Obs: Note that, since $n_0$ is a fixed value $\lim_{n\to\infty}[1-(1-\frac{1}{n})^{n-n_0}] =\lim_{n\to\infty}[1-(1-\frac{1}{n})^{n}]=\frac{1}{e} $
So that $\frac{S_n}{n}\nrightarrow^{P}0$
