Convergence of random variables (Durrett: Probability Theory and Examples)

Its very interesting. Based on Thomas' reply, let me respond:

Let us consider the case where $X_n \geq 0$. Now, we know this property of CCDF $$\lim_{u \to \infty} P(X_n > u) = 0$$

So given $n \geq 1, m \geq 1 \quad\exists \delta_{n,m} > 0$ s.t $$ u \geq \delta_{n,m} \Rightarrow P(X_n > u) \leq \frac{1}{2^m}$$

Now pick $c_{m,n}$ such that $c_{m,n} > \delta_{n,m}2^m$ and substitute above to get $$P\left(\frac{X_n}{c_{m,n}} > \frac{1}{2^m}\right) \leq \frac{1}{2^m}$$

Now arrange the $c_{m,n}$ in a grid and pick the diagonal sequence i.e. let $c_n = c_{n,n}$. Now apply the Borel Cantelli Lemma to conclude:

$\forall \epsilon > 0$, pick m s.t $2^{-m} < \epsilon$ and

$$\sum_{n=1}^{\infty}P\left(\frac{X_n}{c_n} > \epsilon\right) \leq \sum_{n=1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) $$ $$ = \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right)$$ $$ = \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}P\left(\frac{X_n}{c_n} > \frac{1}{2^n}\right)$$

$$ \leq \sum_{n=1}^{m}P\left(\frac{X_n}{c_n} > \frac{1}{2^m}\right) + \sum_{n=m+1}^{\infty}\frac{1}{2^n} < \infty$$ and thus

$$\Rightarrow P\left(\frac{X_n}{c_n}\to 0 \right) =1$$

Hint: Consider $\xi_n(c) := P(|X_n| \geq c)$ and choose $c_n$ such that $\xi_n(2^{-n}c_n)$ is very small.