I'm trying to show the following:
Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of i.i.d random variables with $\mathbb{E}[|X_1|]<\infty$ and $\mathbb{E}[X_1]=\mu$. Consider $$S_n:=X_1+\cdots+X_n,\quad n\in \mathbb{N}$$ Show that if $\mu>0$ or $\mu<0$ then $S_n\to\infty$ resp. $S_n\to -\infty$ almost surely
Now I know that I somehow must argue with the law of large numbers. Here is what I thought. Consider $$W_n=nS_n=n(X_1+\cdots+X_n)$$ Then the law of large numbers tells us that $$p(\lim\limits_{n\to\infty}\frac{1}{n}W_n=\mathbb{E}[nX_1])=1$$ $$p(\lim\limits_{n\to\infty}\frac{1}{n}nS_n=\mathbb{E}[nX_1])=1$$ $$p(\lim\limits_{n\to\infty}S_n=n\mathbb{E}[X_1])=1$$ $$p(\lim\limits_{n\to\infty}S_n=n\mathbb{E}[X_1])=1$$
But here I'm stucked since the limit is only on the left side.
$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers.
If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$.
So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ S_{n}\to+\infty\right\}$$ and consequently $$P(S_n\to+\infty)\geq P(\bar{X}_{n}\to\mu)=1$$