Let $A$ be a square matrix. To show: Matrix exponential converges to some matrix $X$.
$$ \lim_{N \rightarrow \infty} \sum_{k=0}^{N}\frac{A^k}{k!} =X $$
In some proofs that I have seen it is stated that because (for a sub-multiplicative norm) $$ 0 \le \sum_{k=0}^{\infty} \left\Vert \frac{A^k }{k!} \right\Vert \le \sum_{k=0}^{N} \frac{\Vert A \Vert ^k }{k!}<\infty , $$ then the series $\sum_{k=0}^{N}\frac{A^k}{k!}$ has to be convergent. That however isn't clear to me.
To me more intuitive way to show convergence would be to show that $$ \lim_{N \rightarrow \infty} \left\Vert \sum_{k=0}^{N} \frac{A^k}{k!} -X \right\Vert =0$$ and use some intuitive matrix norm for which it is clear that all elements of $\frac{A^k}{k!} -X$ converge to zero.
Any hints?
Consider the largest element of $A$ (in absolute value), let $b$, and consider the matrix $B$ filled with all $b$'s. We have $|A|\le B$ element-wise, then
$$|A^2|\le dbB$$
where $d$ is the order of the matrix, and by induction,
$$|A^n|\le d^{n-1}b^{n-1}B.$$
In other words,
$$|a_{ij}^{(n)}|\le d^{n-1}b^n$$ and $$\frac{|a_{ij}^{(n)}|}{k!}\le\frac{d^{n-1}b^n}{k!}.$$
Hence
$$\left|\sum_{k=0}^\infty\frac{a_{ij}^{(n)}}{k!}\right|\le\sum_{k=0}^\infty\frac{|a_{ij}^{(n)}|}{k!}\le\frac{e^{db}}d$$ converges for every $ij$.