Matrix sequence convergence vs. matrix power series convergence: Is my thinking correct?
The sequence $A^n$ converges if each entry converges to a finite number.
But for a matrix power series, $ I + A + \cdots + A^n + \cdots $ can never converge if it has, for example a "1" in the upper left corner, in entry $a_{11}$.  Take, for simplicty, $A$ to be diagonal, so that the other diagonal entry is $.5$.  This entry will eventually go to zero, but the "$1$" entry will accumulate to infinity.
So we really need, just as in the nth-term test for series convergence of real / complex numbers, for the matrices to tend to the zero-matrix, which I am guessing is a necessary but not sufficient condition for convergence.
What do you think?  
Thanks,
 A: I guess we're talking exclusively about matrix power series here.
It is indeed a necessary (but not sufficient) condition that, in order for $\sum A^n$ to converge, the $i,j$ entry of the matrix power $A^n$ must converge to zero as $n \to \infty$. If 
$$
A = \pmatrix{1\\&0.5\\ &&0.5}
$$
then we indeed find that $\sum A^n$ diverges, since the $1,1$ entry is $1$ for every $n$.
So, if this is what you've been saying, then you're right so far.

However, in practice, it is unnecessarily difficult to compute each $i,j$ entry of a matrix.  In the matter of converging matrices, it is significantly easier to consider a submultiplicative matrix norm. A particularly nice norm of this type is the Frobenius norm.  In particular, we define
$$
\|A\| = \sqrt{\sum_{i=1}^n\sum_{j=1}^n |a_{ij}|^2}
$$
What we can say then is that a necessary condition for the convergence of $\sum A^n$ is that $\|A^n\| \to 0$ as $n \to \infty$.
A more impressive result is that a sufficient condition for the convergence of $\sum A^n$ is that $\|A\| < 1$.
A: It is a necessary condition. By Von Neumann's theorem, we have:
For $A \in M_n(\Bbb C)$, $\sum_{n=0}^{\infty}A^n$ converges if, and only if, $\rho(A) < 1$, where $\rho(A) = \max_{\lambda \in \text{sp}(A)} |\lambda|$.
On the other hand, $\rho(A) < 1 \iff$ there exists an induced (hence multiplicative) matrix norm $\|\cdot \|$ on $M_n(\Bbb C)$, such that $\|A \| < 1$. This implies that: $\|A^n\| \le \| A\|^n \to 0$. Hence, $A^n \to 0$ since all the norms on $M_n(\Bbb C)$ are equivalent.
