Divergence of asymmetric not-simple random walk Consider a (not simple) random walk $S_n = \sum_{k=0}^n X_k$ where X_k are i.i.d and the mean $\overline{X}<0$. Is there is simple proof or a reference showing $P(
\lim \limits_{k \to \infty} S_k = -\infty) = 1$ ?
Also, is there a way to upper bound $P(\bigcup_{k\ge n} (S_k \ge -M) )$ as a function of $M>0$ and $n\ge 0$ (and possibly the first moments of $X_n$). A particular application is the case where $X_k$ can take two values $-a$ and $b$, with $a>b>0$ with probabilities $p$ and $1-p$ respectively with $p>1/2$.
Many thanks in advance, best regards.
 A: Hint on first part of your question preassuming that you want a proof of $S_n\to-\infty$ a.s.:
Note that $\frac1{n}S_n\to\mathbb EX_1<0$ a.s. according to the strong law of large numbers.

edit:
In the sequel I assume that $\mathbb EX_1$ is defined and negative. That's how I understood your "mean $\overline{X}<0$". Correct me if I misunderstood you there. 
If $\mu:=\mathbb EX_1$ then $P(\lim_{n\to\infty}\frac1{n}S_n=\mu)=1$ (application of strong law of large numbers).
So it is enough to prove on base of $\mu<0$ that: $$\lim_{n\to\infty}\frac1{n}s_n=\mu\implies\lim_{n\to\infty}s_n=-\infty$$
Can you do that?
This tells us that $$\{\lim_{n\to\infty}\frac1{n}S_n=\mu\}\subseteq\{\lim_{n\to\infty}S_n=-\infty\}$$
hence$$1=P(\{\lim_{n\to\infty}\frac1{n}S_n=\mu\})\leq P(\{\lim_{n\to\infty}S_n=-\infty\})$$
A: Thanks for the hint. Here is a suggested proof for the first part of the question : 
We have $\lim S_n/n = \overline{X}$ iff
$\forall \epsilon>0, \exists N\in \mathbb{N}, \forall n\ge N, |S_n/n - \overline{X}| < \epsilon $ iff 
$\forall \epsilon>0, \exists N\in \mathbb{N}, \forall n\ge N, |S_n - n\overline{X}| < n\epsilon$ iff 
$\forall \epsilon>0, \exists N\in \mathbb{N}, \forall n\ge N, S_n < \overline{X} n + \epsilon n$.
This implies taking $\epsilon = - \overline{X}/2 >0$ :
$\exists N, \forall n>N, S_n < \overline{X} n - \overline{X}/2 n$ iff
$\exists N, \forall n>N, S_n < \overline{X}/2 n$
implies 
$S_n$ diverges to $-\infty$.
As a consequence $P(S_n \text{ diverges to } -\infty) \ge P(\lim S_n/n = \overline{X}) = 1$ : $S_n \text{ diverges to } -\infty$ almost surely.
