Extension of a well known formula for the Euler number into the complex domain

Question

It is well known that $\lim_{n \to \infty } (1+\frac{r}{n})^{n}= e^{r}$ for every real number $r$, where $e$ denotes the Euler number. For every complex number $z$ a meaning is given to $e^{z}$ (which is then mostly written as $\exp(z)$) by the unique analytic continuation of the real-analytic function $r\mapsto e^{r}$, so it may be defined by the power series $\exp(z)=\sum_{i=0}^{\infty }\frac{z^{n}}{i!}$ which is known to converge absolutely for every complex number $z$. And indeed, $\exp(z)=\lim_{n \to \infty } (1+\frac{z}{n})^{n}$ still holds for every complex number $z$. Can you give a rigorous proof of this?

Answer 1

I have followed up an idea in Conrad's comment to let $n\to \infty$ in the expansion of the binom but there is a sublety, as you cannot pass to the limit of this sum by summing the limits of the single summands because the number of summands is not fixed but $n$ itself, so in general this is not allowed. I could overcome this problem and have worked out a rigorous proof with all details (any comments are welcome):

We define two sequences of functions $\mathbb{C}\to \mathbb{C}$ by $f_{n}(z)= (1+\frac{z}{n})^{n}$ respectively $g_{n}(z)=\sum_{i=0}^{n }\frac{z^{i}}{i!}$. For everything to follow $z\in \mathbb{C}$ is arbitrary but kept fixed. As $\lim_{n \to \infty } g_{n}(z)=e^{z}$ it suffices to show $\lim_{n \to \infty } (g_{n}(z)-f_{n}(z))=0$. Using the binomial theorem $(u+v)^{n}=\sum_{i=0}^{n}\binom{n}{i}u^{i}v^{n-i}$ with $u=\frac{z}{n}$ and $v=1$ we obtain

$f_{n}(z)=(1+\frac{z}{n})^{n}=\sum_{i=0}^{n}\binom{n}{i}(\frac{z}{n}) ^{i}=\sum_{i=0}^{n}\frac{n!}{(n-i)!\cdot n^{i}}\cdot \frac{z^{i}}{i!}$ and thus

(*) $g_{n}(z)-f_{n}(z)=\sum_{i=0}^{n}c_{i,n}\cdot \frac{z^{i}}{i!}$, where $c_{i,n}:=1-\frac{n!}{(n-i)!\cdot n^{i}}$ for integers $n\geqslant i\ge 0$. Let’s have a look at the coefficients $c_{i,n}$:

$c_{0,n}=1-1=0$, $c_{1,n}=1-1=0$, and for integers $n\geqslant i\ge 2$ we have $c_{i,n}=1-(1-\frac{1}{n})\cdot \cdot \cdot (1-\frac{i-1}{n})$

In any case $0\le c_{i,n}\leqslant 1$ and $\lim_{n \to \infty } c_{i,n}=0$ (mind that for $i\geqslant 2$ the number of factors in $c_{i,n}$ is $i-1$ which does not vary with $n$).

Let $R_{m}(r):=\sum_{i=m+1}^{\infty }\cdot \frac{r^{i}}{i!}$ for $r\geqslant 0$ which is meaningful as this series converges for each fixed $r$, and consequently $\lim_{m \to \infty } R_{m}(\left| z \right|)=0$. Thus for arbitrarily chosen $\varepsilon \gt 0$ there is $m_{0} \in \mathbb{N}$ such that $R_{m_{0}}(\left| z \right|)\lt \frac{\varepsilon}{2}$. We now split the sum in (*) at index $m_{0}$ for arbitrary $n\gt m_{0}$:

$g_{n}(z)-f_{n}(z)=S_{n}(z)+T_{n}(z)$, where $S_{n}(z):=\sum_{i=0}^{m_{0}}c_{i,n}\cdot \frac{z^{i}}{i!}$ and $T_{n}(z):=\sum_{i=m_{0}+1}^{n}c_{i,n}\cdot \frac{z^{i}}{i!}$

As $m_{0}$ is fixed while letting $n\to \infty$ we obtain $\lim_{n \to \infty } S_{n}(z)=\sum_{i=0}^{m_{0}} \lim_{n \to \infty } c_{i,n}\cdot \frac{z^{i}}{i!}=0$; therefore $n_{0}\in \mathbb{N}$ can be chosen such that $\left| S_{n}(z) \right|\lt \frac{\varepsilon}{2}$ for all $n\gt n_{0}$. We may well assume $n_{0}\gt m_{0}$

For the second term we estimate $ \left| T_{n}(z) \right|\leqslant \sum_{i=m_{0}+1}^{n} \frac{\left| z \right|^{i}}{i!}\leqslant R_{m_{0}}(\left| z\ \right|)\lt \frac{\varepsilon}{2} $ (independently of $n$) and thus $\left| g_{n}(z)-f_{n}(z) \right|\leqslant \left| S_{n}(z) \right|+\left| T_{n}(z) \right|\lt \varepsilon$ for all $n\gt n_{0}$. As $\varepsilon\gt 0$ had been arbitrary we have shown that $\lim_{n \to \infty } (g_{n}(z)-f_{n}(z)) =0$, q.e.d.