# Show that $\frac{X_{n}}{c_{n}}\rightarrow 0$ almost surely

Suppose $X_{n}$ is any sequence of (real-valued) random variables defined on a common probability space $(\Omega,\mathcal{F},\mathbb{P})$. Show that there exists a sequence $c_{n}\rightarrow \infty$ such that $\frac{X_{n}}{c_{n}}\rightarrow 0$ almost surely. The key is to use the Borel Cantelli lemma, but I am not able to formulate pairwise independent events $A_{n}$ such that $\sum\limits_ {n=1}^{\infty}\mathbb{P}(A_{n})=\infty$ implies $\mathbb{P}(A_{n}~\text{i.o.})=1$, where i.o. stands for infinitely often.

Just a few small hints would be greatly appreciated.

Use (or reprove) the extended form of Borel Cantelli, which says that if there exists a decreasing sequence $\epsilon_i\rightarrow 0$ such that $\sum_i P(|X_i|>\epsilon_i)<\infty$, then $X_i$ converges to 0 almost surely. You can pick any such sequence $\epsilon_i$, such as $\epsilon_n=1/n$, and then pick a corresponding $c_n$ sequence, that decreases $P(|X_n|>c_n/n)$ sufficiently fast.
First note that if $$X_n$$ is real valued then $$$$\mathbb{P}[X_n=\infty] = 0. \hspace{2cm}(1)$$$$ Let $$\{c_n\}$$ be an increasing and unbounded sequence $$(c_n \nearrow \infty)$$. Define $$E_n = \Big \{\omega\in \Omega : |X_n(\omega)| > \frac{c_n}{n}\Big\}$$. Then, in order to get a contradiction, Suppose that there is no sequence of positive numbers $${c_n}$$ such that $$\mathbb{P}[E_n] ≤ 2^{−n},$$ i.e. for every $$N\in\mathbb{N}$$, setting for a fixed $$n$$ $$A_N:= \Big \{\omega\in \Omega : |X_n(\omega)| > \frac{N}{n}\Big\},$$ and $$A_\infty:= \Big \{\omega\in \Omega : |X_n(\omega)| = \infty \Big\},$$
it follows that $$\mathbb{P}[A_{N}] > 2^{-n}.$$ From there, proving that the sequence $$\{A_N\}$$ is decreasing and $$A_\infty = \bigcap_N A_N$$, you can show that $$\mathbb{P}[A_\infty]>0$$ $$\hspace{0.3cm}(\longrightarrow\longleftarrow )$$ since $$(1)$$ implies $$\mathbb{P}[A_\infty]=0$$. Thus, there exists a sequence with characteristics as mention above such that $$\mathbb{P}[E_n] \leq 2^{-n}$$.
So, using comparison criterion for series shows that $$\sum_{n=1}^\infty \mathbb{P}[E_n] < \infty$$, which is all necesary to use Borel - Cantelli lemma to get $$\mathbb{P}[\limsup E_n] = 0$$. This implies that $$0 \leq \frac{|X_n|}{c_n} \leq \frac{1}{n} \longrightarrow 0$$ almost everywhere $$(\omega \notin \limsup E_n)$$ having in mind that $$\mathbb{P}\Big[\Big(\limsup E_n\Big)^c\ \Big] = 1$$, this complete the proof.