Let $A$ a matrix with real or complex entries. Prove that $\displaystyle\lim_{n\rightarrow\infty}\left(E+\frac{A}{n}\right)^n=e^A, E=$identity.
I thought of using the limit, but do not know where converges. $$\displaystyle\lim_{n\rightarrow\infty}\sum_{i=0}^n \dfrac{n!}{(n-i)!n^ii!}$$
Hint: develop $\left(E + \dfrac{A}{n}\right)^n$ using Newton's Binomial and compare with $\displaystyle e^A = \sum \limits_{i=0}^\infty \dfrac{A^i}{i!}$
I assume you are to show point-wise convergence. Let $|B|_{kj}$ denote the modulus of the $(k,j)$ entry of a matrix $B$. Let $2 \leq m \leq n$. Since $E$ commutes with $\frac{1}{n}A$ for any $n$ you can write (following the hint) \begin{align} \left( E + \frac{A}{n} \right)^m &= \sum_{i=0}^m \binom{m}{i}E^{m-i} \left( \frac{A}{n}\right)^i \\ &= \sum_{i=0}^m \binom{m}{i} \left( \frac{A}{n}\right)^i \\ &= \sum_{i=0}^m \frac{A^i}{i!} \frac{m(m-1)\cdots (m-i+1)}{n^i}. \end{align} Since $$ 0 \leq \frac{m\cdots (m-i+1)}{n^i} \leq \frac{n\cdots (n-i+1)}{n^i} \leq 1 $$ when $i \in \{1,\cdots,m\}$, $$ \left| \sum_{i=0}^m \frac{A^i}{i!}\frac{n\cdots (n-i+1)}{n^i} \right|_{kj} \leq \left| \left( E + \frac{A}{n} \right)^n \right|_{kj} \leq \left| \sum_{i=0}^n \frac{A^i}{i!} \right|_{kj} $$ (strictly speaking, the inequality on the left here can only be said to hold when $A$ has non-negative entries, which is too restrictive - I don't see any other way of using the hint though). Take $n \to \infty$ and find $$ \left| \sum_{i=0}^m \frac{A^i}{i!} \right|_{kj} \leq \liminf_{n \to \infty} \left| \left( E + \frac{A}{n} \right)^n \right|_{kj} \leq \limsup_{n \to \infty} \left| \left( E + \frac{A}{n} \right)^n \right|_{kj} \leq \left| e^A \right|_{kj} $$ and then take $m \to \infty$. If you don't use the hint then the approaches suggested in the comments are good.