# Showing that for $X_n$ iid, $\frac{S_n}{n} \to 0$ almost surely implies $\sum_{n=1}^{\infty} P(|X_n| \geq \varepsilon n) < \infty$

I am practicing for an upcoming examination and am having trouble with this suggested exercise. The question is stated as follows:

$$X_1,X_2,\dots$$ are independent and identically distributed on $$(\Omega,\mathcal{F},P)$$ with $$E(X_k)=0$$ $$\forall k$$. Show that if $$\frac{S_n}{n} \rightarrow 0$$ almost surely as $$n\to\infty$$, then $$\sum_{n=0}^{\infty} P(|X_n| \geq \varepsilon n) < \infty$$.

All limits in the solution below are for $$n\to\infty$$:

My idea thus far is that since $$\frac{S_n}{n} = \frac{X_n}{n} + \frac{S_{n-1}}{n-1}\frac{n-1}{n}$$, we can write that $$\frac{X_n}{n} = \frac{S_n}{n}-\frac{S_{n-1}}{n-1}\frac{n-1}{n}$$. Then since both $$\frac{S_n}{n}$$ and $$\frac{S_{n-1}}{n-1}$$ converge almost surely to $$0$$, and since $$\frac{n-1}{n}$$ converges to $$1$$, we have that $$\frac{X_n}{n}\to 0$$ almost surely.

Since almost sure convergence implies convergence in probability, we have $$P \left( \vert \frac{X_n}{n} - 0 \vert > \varepsilon \right) \to 0 \Rightarrow P(|X_n|> \varepsilon n ) \to 0$$.

It remains to show that the summation over this last term is finite. My first instinct is to use part 2 of the Borel-Cantelli lemma which states that for events $$E_n$$, $$P(\limsup E_n)=0 \implies \sum_{n=1}^{\infty} P(E_n) < \infty$$ (using the contrapositive of the statement). If we set $$E_n = \{|X_n| > \varepsilon n\}$$, and we have that $$\lim P(E_n) = 0 \implies \limsup P(E_n) = 0$$. If I could show that $$P(\limsup E_n)=0$$ as well I would be done, but I am unsure of this step. Typically I would use the continuity of the measure $$P$$, but I'm not sure if the events $$E_n = \{|X_n| > \varepsilon n\}$$ are increasing.

I'm also concerned that since I did not use the fact that $$EX_k = 0$$ I have gone down the wrong trail for solving this. Any help would be very much appreciated.

You are almost done. Just use the fact that $$X_i$$s are iid so replace the event $$E_n = \{ |X_n| > \epsilon n\}$$ with $$E_n = \{ |X_1| > \epsilon n \}$$. Now your events are indeed monotone (decreasing).
You don't need the fact that they are mean $$0$$ here, that would be useful for the converse statement.
Edit: Actually, the second Borel-Cantelli you are using requires independence of its events, so, you cannot do the modification above. The result you wanted actually followed directly from the fact that $$X_n/n \rightarrow 0$$ almost surely (no independence needed), since by Borel-Cantelli, that is equivalent to the sum you are trying to compute.