Prove that there exists a numeric sequence $c_1, c_2,...$ such that $\xi_n / c_n \to 0$ almost surely as $n \to \infty$. Let $\xi_1, \xi_2,...$ be a sequence of random variables defined on the same probability space. Prove that there exists a numeric sequence $c_1, c_2,...$ such that $\xi_n / c_n \to 0$ almost surely as $n \to \infty$.
My attempt: I guess the proof needs First Borel-Cantelli Lemma. But I don't know how to prove.
 A: This is just an extension of dmh's answer, which has all the ideas needed to solve the problem. There will be some details in what follows that you should fill in for yourself.
Let $\varepsilon_k>0$ be a sequence such that $\varepsilon_k$ converges monotonically down to $0$. For all $n$ and $k$, there is $m_k(n)\in\mathbb{N}$ big enough so that $\mathbb{P}(|\xi_n| \geq m_k(n)\varepsilon_k)\leq 2^{-n}$. (Why is there such an $m_k(n)$?)  Then we have that
$$
\sum_{n=1}^\infty
\mathbb{P}(|\xi_n| \geq m_k(n)\varepsilon_k)< \infty.
$$
By Borel-Cantelli,
$$
\mathbb{P}(|\xi_n| \geq m_k(n)\varepsilon_k\text{ i.o.}) = 0.
$$
This is equivalent to saying that almost surely there is $N_k\in\mathbb{N}$ such that for all $n\geq N_k$, we have $|\xi_n/m_k(n)| < \varepsilon_k$. (Check this equivalence.)
Now, we need to construct the sequence $(c_n)$. To do this, simply choose $c_n = m_n(N_n)$. Then we almost surely have that $|\xi_n/c_n| < \varepsilon_n$ for all $n$. (Check that this is almost sure.) Thus, for any $\varepsilon>0$, since there is $M\in\mathbb{N}$ such that $\varepsilon_M<\varepsilon$, by construction of $c_n$, we have that $n\geq M$ implies one almost surely has that
$$
|\xi_n/c_n| < \varepsilon_n \leq \varepsilon_M < \varepsilon.
$$
A: Every random variable $X$ is tight so that for every $\epsilon >0$ there exists a $m$ such that:
$$P(X\geq m) \leq \epsilon.$$
So for a given sequence $X_i$ we can choose a sequence $m_i$ such that
$$\sum_i^\infty P(X_i\geq m_i)<\infty.$$
So by the Borel-Cantelli lemma,
$$P(\sup_i X_i\geq m_i) =0.$$
Now choose a sequence $a_i=m_i \epsilon$ and apply the above result to this sequence.
