# Show that, $X_n \rightarrow 0, a.s$ $\Rightarrow$ $S_n/n \rightarrow 0$

Problem: {X_n} is a sequence of random variables (not necessarily independent), $$S_n=\sum_{i=1}^n X_n$$

show that, $$X_n \rightarrow 0, a.s$$ $$\Rightarrow$$ $$S_n/n \rightarrow 0$$

It seems that since $$X_n \rightarrow 0, a.s$$ ,

$$\Rightarrow$$ $$P(|X_n|>\epsilon, i.o)=0$$

That is, $$\lim_{m\rightarrow\infty}P(\bigcup_{n=m}^{\infty}|X_n|>\epsilon)=0$$

We want to show $$P(|S_n/n|>\epsilon, i.o)=0$$

Since, $$P(|S_n/n|>\epsilon, i.o)=\lim_{m\rightarrow\infty}P(\bigcup_{n=m}^{\infty}|S_n/n|>\epsilon)=\lim_{m\rightarrow\infty}P(\bigcup_{n=m}^{\infty}|\sum_{i=1}^nX_i/n|>\epsilon)$$

My question, how to show the last equation converges to $$0$$ as $$n \rightarrow\infty$$ or can anyone give some suggestion of this proof?

• This statement is just true if all $X_i$ are almost sure finite.
– Gono
Commented Dec 16, 2019 at 8:27

Here is a suggestion: show that for any real numbers $$\{a_n\}$$ if $$a_n\to 0$$, then $$\frac{a_1+\dots+a_n}{n}\to 0$$. Now $$X_n\to 0$$ a.s. so we have $$S_n/n\to 0$$ a.s. as well.