Convergence of series of random variable without distribution

Question

I'm trying to solve the following task and I'm struggling very much. I don't know if it is correct what I did so far.

Let $(X_n,n\geq1)$ be a sequence of independent random variables such that $\alpha>0$ and

$$P(X_n=0)=1-\frac{1}{n^{\alpha}}\quad \text{and} \quad P(X_n=n)=\frac{1}{n^{\alpha}},\quad n\geq 1$$

Prove that

1. $X_n\to 0$ as $n\to \infty$ in probability
2. $X_n\to 0$ as $n\to \infty$ almost surely if $\alpha>1$
3. For $r>0,X_n\to0$ as $n\to \infty$ in $L^r$ if $\alpha>r$

My attempt:

For 1 we must show $$\lim\limits_{n\to \infty}P(|X_n-0|>\epsilon)\to 0$$

$$\sum\limits_{n=1}^\infty P(X_n=0) = \sum\limits_{n=1}^\infty 1- \frac{1}{n^\alpha}=\infty$$ So Borel-Cantelli says $$P(\limsup X_n=0) = 1\implies \lim\limits_{n\to \infty}P(|X_n-0|>\epsilon)\to 0 $$

For the second one: If $\alpha >1$ $$\sum\limits_{n=1}^\infty P(X_n=n) = \sum\limits_{n=1}^\infty \frac{1}{n^\alpha}<\infty$$ Then Borell Cantelli tells us that $$P(\limsup X_n=n) = 0$$

Answer 1

Part $1$ is much simpler than what you have. Note that $$P(|X_n| > \epsilon) \leq P(|X_n| > 0) = P(X_n = n) = n^{-\alpha} \to 0.$$

Part $2$ is correct.

For Part $3$, consider the following: $$\|X_n\|_{L^r}^r = n^r \cdot n^{-\alpha} = n^{r - \alpha}.$$ Under what conditions does this go to $0$?