$B(n,z)$ "uniformly" converges to $A(n,z)$ which uniformly converges to G(z)?

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(1). Suppose that $\{A(n,z)\}_{n=2}^{\infty}$ and G(z) are entire functions. Suppose that for $z\in S(1/2):=\{|\mathrm{Im}(z)|< 1/2\}$ and any $0<\epsilon <1$, there exists a positive integer $N=N(\epsilon)$, such that for any integer $n\geqslant N$, we have $\left|G(z)-A(n,z)\right|<\epsilon$. Thus we say that $\{A(n,z)\}_{n=2}^{\infty}$ uniformly converge to $G(z)$ in the critical strip $S(1/2)$.

(2). Suppose that $\{B(n,z)\}_{n=2}^{\infty}$ are entire functions. Supposed that for $z\in S(1/2)$ and any $0<\epsilon <1$, there exists a positive integer $M=M(\epsilon)$, such that for any integer $n\geqslant M$, we have $\left|A(n,z)-B(n,z)\right|<\epsilon$.

Question Do we still say that $\{B(n,z)\}_{n=2}^{\infty}$ uniformly converge to $A(n,z)$ in the critical strip $S(1/2)$?

If not, what kind of name is used for the convergence described in (2)?

The reason I am not quite sure is because in (1) the uniform convergence target function, $G(z)$, is independent of $n$; while as in (2) the convergence target function, $A(n,z)$, is dependent on $n$, so it is a moving target.