30A99-LimitForExpz.tex

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\pmtitle{limit for exp(z)}
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\pmtype{Theorem}
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\begin{document}
For any complex number $z$, we have
\begin{align*}
\lim\limits_{n\to\infty} \left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right)^n=\exp{z},\end{align*} where $ \exp $ denotes the exponential function.
\newline
{\bf Proof:} 
For $\alpha\to 0$, we have 
\begin{align*} 
\ln(1+\alpha) &= \sum_{k=1}^\infty (-1)^{k-1}
\cdot\frac{\alpha^k}{k} \\
&= \alpha+O(\alpha^2). \end{align*}
Therefore 
\begin{eqnarray*}
\left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right)^n &=& \exp\left(n\ln\left(1+\frac{z}{n}+o\left(\frac{1}{n}\right)\right)\right) \\
&=&\exp\left(n\left(\frac{z}{n}+o\left(\frac{1}{n}\right)+O(\frac{1}{n^2})\right)\right)\\
&=&\exp\left(z+o(1)+O(\frac{1}{n})\right)\to\exp{z} \quad \text{for} \quad n\to\infty. \quad \Box \end{eqnarray*}
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