Limit of expectation of product of random variables - using conditional expectation

I'm working on this probability problem and am quite stuck on how to approach the following proof:

$$X_n, n = 1, 2, ... \text{and } Z \text{ are random variables defined on} (\Omega, \mathcal{F}, P)$$

We may assume that:

• $$E|Z| < \infty$$
• $$P(|X_n|\leq c) = 1$$ for some constant $$c > 0$$
• $$EX_n=0$$

The following sigma-algebras are defined as: $$\mathcal{A}_t = \sigma(X_n, n \leq t)$$
$$\mathcal{A}_\infty = \sigma(X_n, n \geq 1)$$

The question asks to prove that if $$Z$$ is $$\mathcal{A}_m$$-measurable then:
$$\lim_{n \rightarrow \infty} E(X_n Z)=0$$

Note we are also given that $$\lim_{t \rightarrow \infty} E(Z|\mathcal{A}_t) = E(Z|\mathcal{A}_\infty) a.s.$$ (1)

I've tried approaching it by substituting $$Z$$ as $$X_nZ$$ into the given limit (1) and then applying dominated convergence theorem but I can't seem to figure out how get the entire expression to converge to 0. Intuitively, it feels like I have to find an equality/inequality which involves $$EX_n$$ but I can't figure out how to get to this result.

I wish I could've progressed further prior to asking for input here but I'm rather new to formally learning probability, so appreciate any advice.

The statement is false. Let $$X$$ have uniform distrbution on $$(-1,1)$$, $$X_n=X=Z$$ for all $$n$$. Then $$EX_nZ=\frac 1 3$$ for all $$n$$.