If $\{A_k\}$ is a sequence of $n \times n \hspace{1mm}$ matrices so that $\sum_{i,j=1}^n|a_{k_{ij}}|^2 < \infty$ for every $k$, i.e sum of squares of all the entries is finite for each matrix of the sequence. How do we ensure or what will be the criterion for convergence of $\exp(A_k)$ to say $\exp(A)$?
We may assume the norm as $$||B||=\sum_{i,j=1}^n|b_{ij}|^2$$ where $B=(b_{ij})_{1\leq i,j \leq n}$. The definition I am using for $\exp(A)$ is $$\exp(A)=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!} +\cdots $$ The above definition of $\exp(A)$ may complicate the proof of convergence hence feel free to share any other definition of $\exp(A)$.
If $A_k\to A$ then $e^{A_k}\to e^A$.
Proof: (In what follows, any norm that satisfies $\|AB\|\le\|A\|\|B\|$ can be used.) $$B^n-A^n=B^{n-1}(B-A)+B^{n-2}(B-A)A+\cdots+B(B-A)A^{n-1}+(B-A)A^{n-1}$$ \begin{align}\therefore\ \|B^n-A^n\|&\le(\|B\|^{n-1}+\|B\|^{n-2}\|A\|+\cdots+\|A\|^{n-1})\|B-A\|\\ &\le {nc}^{n-1}\|B-A\|\end{align} where $c=\max(\|A\|,\|B\|)$.
Turning to the exponential,$$e^B-e^A=(B-A)+\tfrac{1}{2!}(B^2-A^2)+\cdots$$ $$\therefore\ \|e^B-e^A\|\le\|B-A\|(1+\tfrac{2c}{2!}+\tfrac{3c^2}{3!}+\cdots+\tfrac{c^k}{k!}+\cdots)=e^c\|B-A\|$$ So, given any $0<\epsilon<1$, let $k$ be large enough that $\|A_k-A\|<\epsilon$, so $\|A_k\|<\|A\|+\epsilon\le\|A\|+1$; then the above shows $\|e^{A_k}-e^A\|\le e^{\|A\|+1}\epsilon$, which can be made arbitrarily small.