Let $A\in M_n(\mathbb{R})$. Show that $\lim_{n \to \infty} \bigg( I + \frac{A}{n}\bigg)^n = e^A$

Question

I used the binomial theorem and wrote $$\lim_{n \to \infty} \bigg( I + \frac{A}{n} \bigg)^n =\lim_{n \to \infty} \sum_{k=0}^{n} \binom{n}{k} \left(\frac{A}{n} \right)^{k} $$ And so I tried to compare this sum with $$\sum_{i=0}^{\infty} \frac{A^i}{i!}$$ but I couldn't see a clear way to do it.

Any hints?

Answer 1

Hint: prove the result for the complex field. It is true for diagonalizable matrices. Since diagonalizable matrices are dense conclude.