Evaluate $\lim_{j \rightarrow +\infty} (I + A/j)^j$ Let $A$ be a $n\times n$ matrix. Evaluate 
$$
\lim_{j \rightarrow +\infty} \left(I + \frac{A}{j}\right)^j.
$$
My guess is $e^A$.
My attept:  
\begin{align*}
\lim_{j \rightarrow +\infty} (I + \frac{A}{j})^j &= \lim_{j \rightarrow +\infty} \sum_{k=0}^{j} \frac{j!}{k!(j-k)!}I^{j-k}\Big(\frac{A}{j}\Big)^k \\
&= \lim_{j \rightarrow +\infty} \sum_{k=0}^{j} \frac{j!}{k!(j-k)!}\frac{A^k}{j^k}\\
&= \lim_{j \rightarrow +\infty} \sum_{k=0}^{j} \frac{j(j-1)...(j- (k-1))(j-k)!}{k!(j-k)!}\frac{A^k}{j^k} \\
&= \lim_{j \rightarrow +\infty} \sum_{k=0}^{j} \frac{j(j-1)...(j- (k-1))}{k!}\frac{A^k}{j^k}
\end{align*}
Note that $\lim_{j \rightarrow +\infty} \frac{j(j-1)\cdots (j- (k-1))}{j^k} = 1$.
Then $\lim_{j \rightarrow +\infty} (I + \frac{A}{j})^j = e^A$.
How can I prove that I can swap the $\lim$ and $\sum$?
 A: The reason for which you can exchange the limits (i.e. take the limit in $j$ inside) is that the series is uniformly absolutely convergent:
$$
\lim_{j\to \infty}\sum_{k=0}^{j} a_k(j)\quad\textrm{ with }\quad |a_k(j)|\leq \left|\frac{A^k}{k!}\right|\leq \frac{|A|^k}{k!}
$$
With "uniformly" I mean uniformly in $j$: $ |a_k(j)|$ is less than a quantity independent  from $j$.
In practice, the uniform convergence allows you to focus on a finite sum, exchange the limits, make the limit in $j$ and then conclude that the error in taking the finite sum was 'small'. More in detail:
$$
\lim_{j\to \infty}\left(\sum_{k=0}^{N} a_k(j)+\sum_{k=N+1}^{j}a_k(j)\right) =\lim_{j\to \infty}\sum_{k=0}^{N} a_k(j)+\lim_{j\to \infty}\sum_{k=N+1}^{j} a_k(j)=\\
=\sum_{k=0}^{N}\lim_{j\to \infty}a_k(j)+\lim_{j\to \infty} R_N(j)
$$
with $R_N(j):=\sum_{k=N+1}^{j} a_k(j)$. 
Putting $\bar a_k= \lim_{j\to \infty}a_k(j)=A^k/k!$
$$
\lim_{j\to \infty}\sum_{k=0}^{j} a_k(j)
=\sum_{k=0}^{N}\bar a_k+\lim_{j\to \infty} R_N(j)=\sum_{k=0}^{\infty}\bar a_k+\lim_{j\to \infty} (R_N(j)-S_N(j))
$$
 with  $S_N(j):=\sum_{k=N+1}^{j}\bar a_k$. But, since $|a_k(j)|\leq\frac{|A|^k}{k!}$,
$$
\forall j\quad|R_N(j)|\leq \sum_{k=N+1}^{\infty}\frac{|A|^k}{k!}=:r_N,\quad \textrm{with}
\quad \lim_{N\to \infty}r_N=0.
$$
and similarly 
$
\forall j\quad |S_N(j)|\leq\sum_{k=N+1}^{\infty}\frac{|A|^k}{k!}=r_N.
$
PS. When $A$ is a matrix, we define its norm (or one of its norm) as $|A|^2=\sum_{ij}A_{ij}^2 $. The good property (which is used here) is that $|A^n|\leq |A|^n$
