# Show that $n^{\alpha}X_{n} \xrightarrow{n \to \infty} +\infty$ almost surely

Let $$(X_{n})_{n}$$ be a sequence of random variables that are identically distributed on $$\mathcal{U}(0,1)$$. Furthermore, let $$\alpha > 1$$.

Show that $$n^{\alpha}X_{n} \xrightarrow{n \to \infty} +\infty$$ almost surely

My idea: Let $$\epsilon > 0$$

$$\sum_{n \in \mathbb N}P(n^{\alpha}X_{n} \leq \epsilon)=\sum_{n \in \mathbb N}P(X_{n} \leq \frac{\epsilon}{n^{\alpha}})=\sum_{n \in \mathbb N}\frac{\epsilon}{n^{\alpha}}<\infty$$ since $$\alpha > \infty$$

Therefore, by Borel-Cantelli, $$P(\limsup_{n \to \infty}\{n^{\alpha}X_{n} \leq \epsilon\})=0$$

This means, $$\exists N \in \mathbb N$$ so that $$n^{\alpha}X_{n} > \epsilon$$ for all $$n \geq N$$ almost surely BUT $$\epsilon>0$$ was chosen arbitrarily and therefore $$(n^{\alpha}X_{n})_{n\in \mathbb N}$$ is not bounded a.s.

$$\Rightarrow$$ $$n^{\alpha}X_{n} \xrightarrow{n \to \infty} +\infty$$ a.s.

Is my reasoning sound?

I would probably rename $$\epsilon$$ to something else, since ultimately you want to think of it as being a large positive number.
Also, when you write $$\sum_{n \in \mathbb N}P(X_{n} \leq \frac{\epsilon}{n^{\alpha}})=\sum_{n \in \mathbb N}\frac{\epsilon}{n^{\alpha}}$$
This should be $$\le$$ instead of $$=$$, because it is possible for $$\epsilon/n^\alpha$$ to be larger than 1, in which case $$P(X_n \le \epsilon/n^\alpha)$$ is simply 1. But you still get what you want.