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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.

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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.

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