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Let $X_1,X_2,\dots$ be some sequence of random variables.

Prove that there exist some sequence $a_1,a_2,\dots$ of non-zero real numbers such that the sequence $\frac{X_1}{a_1},\frac{X_2}{a_2},\dots$ converges almost surely to $0$.

From Borel–Cantelli only need to show that for every $\epsilon>0$ : $$\sum\limits_{n=1}^{\infty} \Pr\Bigg[\Bigg|\frac{X_n}{a_n}\Bigg|\geq\epsilon\Bigg]<\infty$$

This is true if and only if we can choose $a_1,a_2,\dots$ such that

$$\sum\limits_{n=1}^{\infty} \Pr[|X_n|\geq\epsilon|a_n|]<\infty$$

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Edit (2nd try)

Instead of using a fixed $\epsilon$ in your Borel-Cantelli argument, use something that goes to zero as $n\to \infty,$ like $1/n.$

Since $\lim_{x\to \infty}P(X_n>x)=0$ for all $n$, there is a sequence $(a_n)$ such that $$ P\left(\frac{|X_n|}{a_n} > \frac{1}{n}\right) < \frac{1}{n^2}.$$ Then, BC implies that$$P\left(\frac{|X_n|}{a_n} > \frac{1}{n},\; i.o.\right)=0,$$ which implies $\frac{|X_n|}{a_n} \to 0$ a.s.

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  • $\begingroup$ But why is that the same $a_i$’s for every $\epsilon$? $\endgroup$ Commented Jan 1, 2018 at 13:35
  • $\begingroup$ Yes, you're right (wasn't looking at the big picture). will delete and possibly give it more thought tomorrow. $\endgroup$ Commented Jan 1, 2018 at 14:01
  • $\begingroup$ @TrueTopologist fixed it. $\endgroup$ Commented Jan 1, 2018 at 23:42

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