Evaluate $\lim_{j \rightarrow +\infty} (I + A/j)^j$

Question

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 $\sum$?

Answer 1

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 $|A^n|\leq |A|^n$