# Resnick - Probability Path - Exercise 6.16 (a)

I'm trying to solve the following exercise from Resnick's book:

For any sequence of random variables {$X_n$} set $S_n=\sum_{i=1}^{n}X_i$

Show $X_n \xrightarrow{a.s.} 0$ implies $S_n/n \xrightarrow{a.s.} 0.$

I know that if $X_n$ converges almost surely, then its Cesaro sum must also converge. But how to formalize the proof?

Note that I can't apply Strong Law of Large Numbers, since it applies for any sequence, so that the theorem may not be applicable.

This can be solved pointwise.

Let $(x_n)_n$ be a sequence in $\mathbb R$ with $\lim_{n\to\infty}x_n=0$, let $s_n:=\sum_{i=1}^n x_i$ and let $\bar x_n:=s_n/n$.

$\lim_{n\to\infty}x_n=0$ implies the existence of some $c>0$ such that $|x_n|\leq c$ for every $n\in\mathbb N$.

If $\epsilon>0$ then some $N\in\mathbb N$ exists with $n>N\implies |x_n|<\epsilon$.

Then for $n>N$ we have: $$|\bar x_n|\leq\frac1n\sum_{i=1}^n|x_n|\leq\frac{N}{n}c+\frac{n-N}{n}\epsilon\leq \frac{N}{n}c+\epsilon$$

Then consequently $$\limsup|\bar x_n|\leq\epsilon$$

This for every $\epsilon>0$ so$$\limsup|\bar x_n|=0\text{ or equivalently }\lim_{n\to\infty}\bar x_n=0$$

This proves that: $$\{X_n\to0\}\subseteq\{S_n/n\to0\}$$ so that $$\mathsf P(X_n\to0)=1\implies\mathsf P(S_n/n\to0)=1$$