TITLE: Prove that $\displaystyle\lim_{k \to \infty} \left( I + \frac{1}{k}A \right)^{k} = e^A$
QUESTION [12 upvotes]: I'm having a little trouble here to prove the following statement:

Let $A$ be an $n \times n$ matrix (real or complex). Prove that
$$\lim_{k \to \infty} \left( I + \frac{1}{k} A \right)^{k} = e^{A}$$

Now I'm using matrix and possible non-commutative; I don't know where to begin. Can you give a spit? Thanks for your attention!

REPLY [8 votes]: Here is an answer without using diagonal arguments.
Let $\| \cdot \|$ be a sub-multiplicative norm over $M_n(\mathbb{C})$ meaning
$$\forall (A,B) \in M_n(\mathbb{C})^2, \quad \|AB\|\leq\|A\|\space\|B\|$$
(The Schatten p norm is an exemple of such a norm)
We can easily prove:
$$(\text {Lemma 1}): \forall A\in M_n(\mathbb{C}),\space \forall k \in \mathbb{N}\space\|A^k\|\leq\|A\|^k$$
Let $A \in M_n(\mathbb C)$
$$\eqalign{\left\| \left( I_n + \frac A p \right)^p - \exp A \right\| &
= 
\left\| \sum\limits_{k=0}^p {\binom  p k} {\frac {A^k} {p^k}} - \sum\limits_{k=0}^\infty \frac{A^k}{k!} \right\|\\
&= 
\left\| \sum\limits_{k=0}^p \frac {A^k} {k!} \left( \frac {p!} {(p-k)!p^k}-1\right) - \sum\limits_{k=p+1}^\infty \frac{A^k}{k!}
\right\|\\
&\leq
\sum\limits_{k=0}^p  {\frac {\left\| {A^k} \right\|}  {k!}} \left| \frac {p!} {(p-k)!p^k}-1 \right|+ \sum\limits_{k=p+1}^\infty {\frac {\left\| {A^k} \right\|}{k!}}&(\text{Triangular Inequality})\\
&\leq
\sum\limits_{k=0}^p  {\frac {\left\| {A} \right\|^k}  {k!}} \left(1- \frac {p!} {(p-k)!p^k} \right)+ \sum\limits_{k=p+1}^\infty {\frac {\left\| {A} \right\|^k}{k!}} & (\text {Lemma 1})\\
&=
\sum\limits_{k=0}^\infty  {\frac {\left\| {A} \right\|^k}  {k!}} - \sum\limits_{k=0}^p {\frac {\left\| {A} \right\|^k}{k!}}\frac {p!} {(p-k)!p^k}\\ 
&=
\exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p
}$$
We proved:
$$0\leq\left\| \left( I_n + \frac A p \right)^p - \exp A \right\|
\leq
\exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p$$
But $$\exp \left\| {A} \right\|-\left( 1+\frac {\left\| {A} \right\|}{p}\right)^p \underset {p\to {\infty}} {\longrightarrow} 0 $$
then by the squeeze theorem:
$$\left\| \left( I_n + \frac A p \right)^p - \exp A \right\|\underset {p\to {\infty}} {\longrightarrow} 0 $$
which is equivalent to:
$$  \left( I_n + \frac A p \right)^p \underset {p\to {\infty}} {\longrightarrow} \exp A $$