Show $\liminf \frac{S_n}{n} = -\infty$ for a sequence of iid random variables I have a problem with solving this question.

$(X_n)$ is i.i.d sequence with $E|X_1| = \infty$ and $\limsup_n \frac{S_n}{n} < \infty$ with positive probability, where $S_n = X_1 + \ldots+ X_n$. Then, prove that $\liminf_n \frac{S_n}{n} = -\infty$ almost surely.

At first, I tried several times including the similar process of proof of Thm 2.4.5 in Durrett, which states that if $X_n$ is i.i.d and $EX_i^{+} = \infty$ and $EX_i^{-} < \infty$, $\frac{S_n}{n} \rightarrow \infty$ a.s. But, I think that none of the proofs work well. 
And I don't have a brilliant way to solve this, please help me.
 A: Hints:


*

*Show that $$\limsup_{n \to \infty} \frac{|X_n|}{n} = \infty \quad \text{a.s.}$$ Hint: Using $\mathbb{E}(|X_1|) = \infty$ show that $$ \sum_{n \geq 1} \mathbb{P}(|X_n| \geq nk) = \sum_{n \geq 1} \mathbb{P}(|X_1| \geq nk) = \infty$$ for any $k \geq 1$. Apply the Borel-Cantelli lemma.

*Recall the following elementary result:

If $\limsup_{n \to \infty} |a_1+\ldots+a_n|/n < \infty$, then $\limsup_{n \to \infty} |a_n|/n< \infty$.

Use this statement to deduce from step 1 that $$\limsup_{n \to \infty} \frac{|X_1+\ldots+X_n|}{n} = \infty \quad \text{a.s.} \tag{1}$$

*Define $$\begin{align*} A &:= \left\{ \limsup_{n \to \infty} \frac{S_n}{n} = \infty \right\} \qquad B := \left\{ \liminf_{n \to \infty} \frac{S_n}{n} = -\infty \right\}. \end{align*}$$ Apply a 0-1-law of your choice to show that $\mathbb{P}(A) \in \{0,1\}$ and $\mathbb{P}(B) \in \{0,1\}$.

*By assumption, $\limsup_n S_n/n< \infty$ with positive probability, and therefore, by step 3, $\mathbb{P}(A)=0$.

*Conclude that $$\mathbb{P} \left( \liminf_{n \to \infty} \frac{S_n}{n} = - \infty \right) = \mathbb{P} \left( \limsup_{n \to \infty} \frac{|S_n|}{n} \right) \stackrel{(1)}{=} 1.$$

