Meaning of $\lim\limits_{n\rightarrow \infty}\mathbb{E}(X_n^2)=0$? This is related to convergence in probability:
What's the meaning of $\lim\limits_{n\rightarrow \infty}\mathbb{E}(X_n^2)=0$ for random variables $X_1,X_2,...$.
 A: The statement is straight forward in terms of definitions. As hinted in the comments, for every $n$, $\mathbb E X_n^2$ is a real number, with the caveat that it may also be positive or negative infinity. This caveat is unimportant if you're told that $\mathbb EX_n^2 \to 0$, however, since then only finitely many of the terms may be infinite and you can ignore those when considering the limit. Indeed, by the definition of limits of sequences of real numbers the statement says that for every $\epsilon >0$ there exists some constant $M_\epsilon$ such that $\mathbb EX_n^2 < \epsilon$ for every $n\geq M_\epsilon$.
The statement in question implies that the sequence of random variables $X_n, n=1,2,\dots$ converges in probability to zero. Here is a proof that $\mathbb E \vert X_n\vert \to 0$ implies convergence in probability to zero; I leave it to you to fill in the rest.
Proof. We work on a probability space $(\Omega, \mathcal F, P)$ so that $\mathbb E$ denotes integration with respect to the measure $P$. Suppose $X_n$ does not converge in probability to zero, that is, there exists $\epsilon>0$ and $\delta >0 $ such that $P(\vert X_n \vert > \epsilon)>\delta$ for infinitely many $n$. Then for every such $n$, $\mathbb E \vert X_{n} \vert \geq \epsilon\delta$. Since there are infinitely many such $n$, there does not exist a constant $M_{\epsilon\delta}$ as defined above and this finishes the proof.
