Let $(X_n)_{n \in \mathbb{N}}$ a series of random variables with:
$$P(X_n = 2^n) = \frac{1}{2^n} \hspace{15pt}\text{and}\hspace{15pt} P(X_n = 0) = 1-\frac{1}{2^n}$$
for all $n \in \mathbb{N}$.
Question: Does a random variable $Z$ exist with $X_n \rightarrow Z$ (in distribution), $X_n \rightarrow Z$ (in probability), $X_n \rightarrow Z$ (in $\mathcal{L}_2)$?
(my) Solution: I saw that $X_n \rightarrow 0$ (in probability) and proved it. For $\epsilon > 0$ we get:
$$P(|X_n - 0| > \epsilon) = P(|X_n| > \epsilon) \leq P(|X_n| \neq 0) \\ = 1 - P(|X_n| = 0) = \frac{1}{2^n} \rightarrow 0 \text{ as } n \rightarrow \infty$$ Now this already implies that $X_n \rightarrow 0$ (in distribution) and that if there exists a random variable $Z$ with $X_n \rightarrow Z$ (in $\mathcal{L}_2)$ $Z \equiv 0$. I proceeded by testing:
$$||X_n - Z||_{\mathcal{L_2}}^2 = E(|X_n|^2) = (2^n)^2\cdot\frac{1}{2^n} + 0\cdot (1 - \frac{1}{2^n}) = 2^n \neq 0 \text{ as } n \rightarrow \infty$$
And concluded that $(X_n)_{n \in \mathbb{N}}$ does not converge in $\mathcal{L}_2$.
Is this argumentation correct? Is there a way to approach problems like this more efficiently (I just thought that $X_n \rightarrow 0$ (in probability) and tested it)?
I agree for the reasoning for the convergence in probability.
For the $\mathbb L^2$-convergence, you used implicitely the fact that if a sequence converges in $\mathbb L^2$ to $Z$ and in probability to $Z'$ then $Z=Z'$ almost surely, which is true. It rests on the fact that convergence in $\mathbb L^2$ implies convergence in probability.
As pointed out in the comment, the sequence $\left(X_n\right)_{n\geqslant 1}$ is also convergent to $0$ almost surely. To see this, one case use the Borel-Cantelli lemma with $A_n:=\{X_n\neq 0\}$.