Convergence of exponential matrix sum 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?)
 A: This is just $\exp(tA)-I$, the matrix exponential, except that you forgot to start summation at$~0$. The sum always converges for the same reason the ordinary exponential does, namely the $k!$ in the denominator beats the at most exponential growth of each entry of $(tA)^k$. To be precise, each entry of $(tA)^k$ is bounded in absolute value by $(nm)^k$ where $m$ is the maximal absolute value of all entries of $tA$; this can be shown by an easy induction on $k$.
A: Treat $A$ as a linear transformation $\mathbb R^n \to \mathbb R^n$. As it has an operator norm and $||A^k|| \leq ||A||^k$. Then the series 
$$B = \sum_{k=0}^\infty \frac{(At)^k}{k!}$$ 
converges absolutely, as 
$$\sum_{k=0}^\infty \bigg|\bigg| \frac{(At)^k}{k!}\bigg|\bigg|\leq \sum_{k=0}^\infty \frac{(||A||t)^k}{k!} = e^{||A||t}$$
As the space of linear transformation is complete, the series converges for all $t$.
A: 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.
A: If $A$ is diagonalizable, $A=P^{-1}DP$, with $D$ a diagonal matrix whose entries are $\lambda_1,...,\lambda_n$, then the sum is $$P^{-1}\left(\sum_{k=1}^{\infty}\frac{D^kt^k}{k!}\right)P$$ and the matrix in the middle is a diagonal matrix whose entries are easy to calculate.
I don't know how to do it if $A$ is not diagonalizable.
