TITLE: Convergence of exponential matrix sum
QUESTION [6 upvotes]: Let $A$ be an $n\times n$ matrix. Consider the infinite sum $$B=\sum_{k=1}^\infty\frac{A^kt^k}{k!}$$ Each term $\dfrac{A^kt^k}{k!}$ is an $n\times n$ matrix. Does the sum $B$ always converge? (i.e. does the sum for each of the $n^2$ entries always converge?)

REPLY [2 votes]: Note that the Frobenius norm $\|A\|_F = \sqrt{\sum_{i,j} |[A]_{ij}|^2}$ is submultplicative.
Then $\|A^k\|_F \le \|A\|_F ^k$. And so $\|e^A\|_F = \|\sum_k \frac{1}{k!} A^k \|_F \le \sum_k \frac{1}{k!} \|A\|_F^k = e^{\|A\|_F}$. In particular, since $|[A]_{ij}| \le \|A\|_F$, we see that $|[e^A]_{ij}| \le e^{\|A\|_F}$, hence each entry converges.