Suppose $A \in M_{d \times d}(\mathbb{C})$. The issue of the convergence of the series of the exponential function of $A$ has been already addressed in this site. Some nice answers can be found here which clearly explains how to use sub-multiplicative norm (more specifically operator norm) to show that $e^A \to \sum_{n=1}^{\infty}\frac{A^n}{n!}$. The striking feature of the usage of sub-multiplicative norm is that it allows us to claim that $||A^n||\le ||A||^n$. Now it is not hard to prove the above convergence as the series $\sum_{n=1}^{\infty}\frac{x^n}{n!}$ converges for all $x \in \mathbb{R}$
But is there not any possibility to reach the same conclusion without using a sub-multiplicative norm?
What I attempted:- Let $A=\left[a_{ij}\right]_{d \times d}$. I would try to use Max norm which is defined as $||A||_{\max}=\max_{ij}|a_{ij}|$ in order to reach the conclusion. I will write $||A||$, in place of $||A||_{\max}$, to denote the max norm of $A$.
In one of my recent question, which can be found here, I made an attempt to show that for any $A \in M_{d \times d}(\mathbb{C})$ \begin{equation} ||A^n||\le d^{n-1}||A||^n \end{equation} Since $d$ is positive, it is legitimate to write \begin{equation}
||A^n||\le d^{n-1}||A||^n<d^n ||A||^n\dots \dots (1)
\end{equation}
Let \begin{equation} \begin{aligned} s_n &=||I||+||A||+\frac{||A^2||}{2!}+\frac{||A^3||}{3!}+\dots+\frac{||A^n||}{n!}\\
&=1+||A||+\frac{||A^2||}{2!}+\frac{||A^3||}{3!}+\dots+\frac{||A^n||}{n!}\\
&\le 1+d||A||+\frac{\left(d||A||\right)^2}{2!}+\frac{\left(d||A||\right)^3}{3!}+\dots+\frac{\left(d||A||\right)^n}{n!}\quad \left(\mbox{From $(1)$}\right)
\end{aligned}
\end{equation}
The right hand side of the above inequality tends to $e^{d||A||}$ as $n \to \infty$. Thus, \begin{equation}
\lim_{n\to \infty}s_n \le e^{d||A||} < \infty
\end{equation}
The immediate consequence of the above result is that the series $\sum_{n=1}^{\infty} \frac{A^n}{n!}$ converges.
Am I correct? I would highly appreciate any valuable guidance provided by the scholars of this mathematical community.
