Let $(X_n)_{n\in \mathbb{N}}$ be a sequence of independent random variables with $$\Pr(X_n=2^n)=2^{-n},\quad \Pr(X_n=0)=1-2^{-n}$$ Show $ X_n\to^\Pr 0$ and $X_n\to 0$ almost surely.
I have shown: $$EX_n=\sum_{k} k \Pr(X_n=k)=2^n\Pr(X_n=2^n)+0=2^n\cdot 2^{-2}=1$$
$$1=EX_i=\int_0^\infty\Pr(|X_i|>t)\ dt=\sum_{n=0}^\infty \int_n^{n+1}\Pr(|X_i|)\leq \sum_{n=0}^\infty\Pr(|X_i| \geq n) $$
To show $X_n\to^\Pr 0$, we have to show $$\forall \epsilon >0: \lim_{n\to \infty} \Pr(\{|X_n|>\epsilon\})=0$$ I am not sure how to show this.
For any $\varepsilon >0$, note that $$ 1-2^{-n}=P(X_{n}=0)\leq P(|X_{n}|\leq \varepsilon)\leq 1 $$ Let $n\to \infty$ to get the result.