Use the fact that for all e>0, P(|S_n/n|>e i.o.)=0 implies P(|S_n/n| does not converge to 0)=0.

To prove P(|S_n/n|>e i.o.)=0, we use Borel-Cantelli and only need to prove

/sum_{n=1}^{Infinity} P(|S_n/n|>e) < infinity.

To prove the above, we use Chernoff bound

P(|S_n/n|>e)<=exp(-nI(e))

and the fact that I(e)>0(?).

What’s the difference between the above and Komogolov‘s SLLN?

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