For this question, it is defined that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$, $\forall x\in\mathbb{C}$.

Though I know why $\lim_{n\rightarrow \infty}(1+x/n)^n=e^x$ pointwise for any real $x$, I feel difficult to show that the sequence of functions converges uniformly on each bounded subset in $\mathbb{C}$. This question comes up in an undergraduate analysis class, but I have little experience working with the complex number. Could you help me on that? Thank you!


Consider the power series for $\log$ which converges in the complex plane if $|z| < 1:$

$$\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \ldots.$$

Hence, if $|z| < n$ we have

$$\log(1+z/n) = \frac{z}{n} - \frac{z^2}{2n^2} + \frac{z^3}{3n^3} - \frac{z^4}{4n^4} + \ldots,$$

and, using the triangle inequality

$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{2n} + \frac{|z|^3}{3n^2} + \frac{|z|^4}{4n^3} + \ldots.$$

In a bounded subset $S \subset \mathbb{C}$, we have a positive real number $R$ such that $|z| \leqslant R$. For all sufficiently large $n$, we have $n > 2R \geqslant 2|z| \implies |z|/n \leqslant 1/2$ for all $z \in S.$


$$\left|n\log(1+z/n) - z\right| \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{|z|}{3n} + \frac{|z|^2}{4n^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 2^2} + \ldots\right] \\ \leqslant \frac{|z|^2}{n}\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right] \\ = \frac{|z|^2}{n} \\ \leqslant \frac{R^2}{n}.$$


$$\lim_{n \to \infty} n\log(1+z/n) = \lim_{n \to \infty} \log(1+z/n)^n = z,$$

uniformly for $z \in S$.

The function $z \mapsto \exp(z)$ is uniformly continuous on the compact set $\{z : |z| \leqslant R \},$ and it follows that

$$\lim_{n \to \infty} (1+z/n)^n = e^z, $$

uniformly for $z \in S$.


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