limit for exp(z)

For any complex number $z$, we have

$\underset{n\to \mathrm{\infty }}{lim}{\left(1+{\displaystyle \frac{z}{n}}+o\left({\displaystyle \frac{1}{n}}\right)\right)}^n=\exp z,$

where $\exp$ denotes the exponential function.
Proof: For $\alpha \to 0$, we have

$\ln (1+\alpha )$ $={\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{(-1)}^{k-1}\cdot {\displaystyle \frac{{\alpha }^k}{k}}$

$=\alpha +O({\alpha }^2).$

Therefore

${\left(1+{\displaystyle \frac{z}{n}}+o\left({\displaystyle \frac{1}{n}}\right)\right)}^n$ $=$ $\exp \left(n\ln \left(1+{\displaystyle \frac{z}{n}}+o\left({\displaystyle \frac{1}{n}}\right)\right)\right)$

$=$ $\exp \left(n\left({\displaystyle \frac{z}{n}}+o\left({\displaystyle \frac{1}{n}}\right)+O({\displaystyle \frac{1}{n^2}})\right)\right)$

$=$ $\exp (z+o(1)+O({\displaystyle \frac{1}{n}}))\to \exp z\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}n\to \mathrm{\infty }.\mathit{\hspace{1em}}\mathrm{\square }$

Title	limit for exp(z)
Canonical name	LimitForExpz
Date of creation	2013-03-22 14:34:54
Last modified on	2013-03-22 14:34:54
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	11
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 30A99