Given
$$\displaystyle \mathbb P(X_n=1)=\frac{1}{n^{\alpha}},~~~\text{and}~~~\displaystyle \mathbb P(X_n=0)=1-\frac{1}{n^{\alpha}}$$ where in $\alpha>0$, $n\geq1$.
How can I prove that $$X_n\rightarrow 0~~\text{in}~~~ \mathcal L^{1}$$
Also, how do I show that $$\displaystyle \limsup_{n\rightarrow +\infty}X_n= \begin{cases} 1, & \text{if }\alpha \leq 1 \\ 0, & \text{if }\alpha >1 \end{cases}$$
Thanks for any help.
Attempt: The sequence converges in probability to $0$ if and only if $\displaystyle \lim_{n\rightarrow \infty}\left(1-\frac{1}{n^{\alpha}}\right)=0$.
By Borel-Cantelli Lemma the sequence converges almost surely to 0 if and only if
$\displaystyle \sum_{n=1}^{\infty}\left(1-\frac{1}{n^{\alpha}}\right)<\infty$
