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Unformatted text preview: ω > 0 then Y n ( ω ) = 0 for all n such that p n < ω. Thus Y n ( ω ) → 0 for all ω ∈ (0 , 1] , which is a probability 1 subset. The second Borel  Cantelli lemma fails because { Y n } are not independent: P { Y n = 1 ,Y n +1 = 1 } = p n +1 6 = p n p n +1 as long as 0 < p n ,p n +1 < 1 . Problem 4 Apply the second BC lemma: For any C < ∞ , deﬁne j ( n ) = log 2 ([ Cn log 2 n ]) . Then P ( X n > Cn log 2 n ) = X j :2 j >Cn log 2 n 2j = 2j ( n ) ≥ 1 Cn log 2 n = ∞ Therefore, P ( X n > Cn log 2 n i.o. ) = 1 . Thus, S n n log 2 n ≥ X n n log 2 n > C i.o. and since C was arbitrary we can make it arbitrarily large and we have the conclusion....
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 Spring '09
 Reed
 Trigraph, yn, Sn, Arbitrarily large, 2J

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