Why is $e^{iAt/n} = I + \frac{1}{n} i A t + O(\frac{1}{n^2})$?

Question

Currently, I am reading Section 4.7.2 of Quantum Computation and Quantum Information: 10th Anniversary Edition by Nielson et. al.

In the proof of Theorem 4.3, the authors have mentioned that:

$$e^{iAt/n} = I + \frac{1}{n} i A t + O(\frac{1}{n^2})$$

where, $A$ is a Hermitian operator and $n$ is a positive integer.

I would like to understand why. Thanks.

Answer 1

$e^{sB}=I+sB+s^2( \frac{B^2}{2!}+s \frac{B^3}{3!}+...)$ and $\frac{B^2}{2!}+s \frac{B^3}{3!}+... \to \frac{B^2}{2!}$ for $s \to 0$.