Proving the Lie-Product formula Let $A,B\in gl(n)$. Then prove
$$e^{A+B} = \lim_{k\to\infty} \left (e^{\frac{A}{k}}e^{\frac{B}{k}}\right )^k$$
I found this theorem in this notes http://www4.ncsu.edu/~aalexan3/articles/liegroups.pdf but there is no proof. I am new to exponential of matrices and can't prove this by my own.
 A: Let $A,B \in \mathcal{B}(H)$ be bounded operators on some Banach space $H$. 
Define $C := e^{(A+B)/k}$ and $D := e^{A/k}e^{B/k}$. We have the following estimates for the norm of $C,D$:
$$\|C\|, \|D\| \leq \exp\left[\frac{\|A\| + \|B\|}{k}\right] = \exp(\|A\| + \|B\|)^{1/k} $$
Applying the Cauchy-Product formula on $D$ yields
$$ D = e^{A/k}e^{B/k} = \sum_{i=0}^{\infty} \frac{(A/k)^{i}}{i!} \cdot \sum_{j=0}^{\infty} \frac{(B/k)^{j}}{j!} = \sum_{m=0}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!}$$ 
Which permits us to bound the norm of $C-D$ by
\begin{align*}
 \|C-D\| &= \left\|\sum_{i=0}^{\infty} \frac{([A+B]/k)^{i}}{i!} - \sum_{m=0}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!} \right\| \\ 
 &= \left\|\sum_{i=2}^{\infty} k^{-i} \frac{(A+B)^{i}}{i!} - \sum_{m=2}^{\infty} k^{-m} \sum_{i=0}^{m} \frac{A^{i}}{i!} \cdot \frac{B^{m-i}}{(m-i)!} \right\| \\
 &\leq \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \sum_{i=0}^{m} \frac{\|A\|^{i}}{i!} \cdot \frac{\|B\|^{m-i}}{(m-i)!}\right] \\
&= \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \frac{1}{m!}\sum_{i=0}^{m} \binom{m}{i}\|A\|^{i} \cdot \|B\|^{m-i}\right]\\
&= \frac{1}{k^2} \cdot\left[ \exp(\|A\| +\|B\|) + \sum_{m=2}^{\infty} \frac{(\|A\|+\|B\|)^m}{m!}\right] \\
 &\leq \frac{2}{k^2} \cdot \exp(\|A\| +\|B\|) 
\end{align*}
We obtain $$\| C^{k} - D^{k} \| = \left\| \sum_{m=0}^{k-1} C^{m} (C-D) D^{k-m-1}\right\| \leq \exp(\|A\|+\|B\|)\cdot k \cdot \|C-D\|$$
where I used that $\|C\|^m\cdot \|D\|^{k-m-1} \leq \exp(\|A\|+\|B\|)^{\frac{k-1}{k}} \leq \exp(\|A\|+\|B\|).$ Together with the estimate for $\|C-D\|$, this yields $$\| C^{k} - D^{k} \| \leq \frac{2\exp(2\|A\| + 2\|B\|)}{k} $$ which proves the claim.
A: For any matrix $P$,
$$
e^P = I + P + {P^2 \over 2!} + {P^3 \over 3!} + \cdots + {P^n \over n!} + \cdots
$$
Thus, it follows that
$$
e^{A \over m} e^{B \over m} = I + {A \over m} + {B \over m} + O\left({1 \over m^2} \right)
$$
Since $e^{A \over m} e^{B \over m} \to I$ as $m \to \infty$, we see that
$e^{A \over m} e^{B \over m}$ is in the domain of the logarithm for all $m$ sufficiently large.
We also know that for all $n \times n$ matrices with $\Vert P \Vert < {1 \over 2}$,
$$
\log(I + P) = P + O(\Vert P \Vert^2)
$$
Hence, for large values of $m$, we have
$$
\log\left(  e^{A \over m} e^{B \over m} \right) = \log\left( I + {A \over m} + {B \over m} + O\left({1 \over m^2} \right) \right)
$$
i.e.
$$
\log\left( e^{A \over m} e^{B \over m}  \right)  =  {A \over m} + {B \over m} + O\left( 
\left\Vert {A \over m} + {B \over m} + O\left({1 \over m^2}\right) \right\Vert^2 \right) 
$$
Thus,
$$
\log\left( e^{A \over m} e^{B \over m}  \right)  = {A \over m} + {B \over m} + O\left( {1 \over m^2} \right)
$$
Exponentiating the logarithm, we get
$$
e^{A \over m} e^{B \over m} = \exp\left[  {A \over m} + {B \over m} + O\left( {1 \over m^2} \right) \right]
$$
Therefore,
$$
\left( e^{A \over m} e^{B \over m} \right)^m =
 \exp\left[  A + B + O\left( {1 \over m} \right) \right]
$$
By the continuity of the exponential, we conclude that
$$
\lim\limits_{m \to \infty} \ \left( e^{A \over m} e^{B \over m} \right)^m
= exp(A + B)
$$
which proves the Lie product formula.
