# limit for exp(z)

For any complex number $z$, we have

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

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

 $\displaystyle\ln(1+\alpha)$ $\displaystyle=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot\frac{\alpha^{k}}{k}$ $\displaystyle=\alpha+O(\alpha^{2}).$

Therefore

 $\displaystyle\left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right)^{n}$ $\displaystyle=$ $\displaystyle\exp\left(n\ln\left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right% )\right)$ $\displaystyle=$ $\displaystyle\exp\left(n\left(\frac{z}{n}+o\left(\frac{1}{n}\right)+O(\frac{1}% {n^{2}})\right)\right)$ $\displaystyle=$ $\displaystyle\exp\left(z+o(1)+O(\frac{1}{n})\right)\to\exp{z}\quad\text{for}% \quad n\to\infty.\quad\Box$
Title limit for exp(z) LimitForExpz 2013-03-22 14:34:54 2013-03-22 14:34:54 mathcam (2727) mathcam (2727) 11 mathcam (2727) Theorem msc 30A99