Consider a sequence of random variables $\{M_n\}$, a random variable $Z$, and the sequences of real numbers $\{a_n\}$, $\{b_n\}$.
Suppose that $Z_{n}\equiv \frac{M_n-a_n}{b_n}\Rightarrow_n Z$, where $\Rightarrow_n$ means weak convergence as $n\rightarrow \infty$, i.e.
$$ \lim_{n\rightarrow \infty} P(\frac{M_n-a_n}{b_n}\leq x)=P(Z\leq x) \text{ }\forall \text{ points of continuity of the cdf of $Z$} $$
Does this imply that, $\forall m \in \mathbb{N}$, $Z_{mn}\equiv \frac{M_{mn}-a_{mn}}{b_{mn}}\Rightarrow_n Z$?
My attempt: using the limit definition, fixing $x\in \mathbb{R}$, and $m=3$ $$ \lim_{n\rightarrow \infty} P(\frac{M_n-a_n}{b_n}\leq x)=P(Z\leq x) \text{ }\forall \text{ points of continuity of the cdf of $Z$} $$ $$ \Updownarrow $$ $$ \forall \epsilon>0 \text{ }\exists \bar{n}_{\epsilon} \text{ s.t. } \forall n\geq \bar{n}_{\epsilon} \text{ } |P(\frac{M_n-a_n}{b_n}\leq x)-P(Z\leq x)|\leq \epsilon $$ $$ \Downarrow $$ $$ \forall \epsilon>0 \text{, }\forall \bar{n}_{\epsilon} \text{ found above, } \forall n\geq \bar{n}_{\epsilon} \text{ } |P(\frac{M_{3n}-a_{3n}}{b_{3n}}\leq x)-P(Z\leq x)|\leq \epsilon $$ How should I conclude?
Let $x$ be a continuity point of the cumulative distribution function of $Z$. Define $c_n:=\mathbb P\left(\frac{M_n-a_n}{b_n}\leqslant x\right)$. We know that $\left(c_n\right)_{n\geqslant 1}$ converges to $\mathbb P\left(Z\leqslant x\right)$ hence so does the sequence $\left(c_{mn}\right)_{n\geqslant 1}$.