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

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?

• there are many ways to approach this and even prove the $e^z$ result from scratch without assuming it on the reals - show that the limit on the LHS is uniform in $z \in K$ compact and hence it can be differentiated term by term so you get that if the limit is $g(z)$ then $g'(z)=g(z), g(0)=1$ so $g(z)=e^z$; in the same vein, write $g(z)=\sum a_kz^k$ and is immediate that $a_k=1/k!$ by expanding the binomial for large $n$ and taking $n \to \infty$; if you assume it on reals, use analytic continuation, the limit is entire, but it is $e^r$ on the reals so it is $e^z$ etc Commented Feb 26, 2023 at 18:34
• This has been done on MSE before... Commented Feb 28, 2023 at 16:53
• Let me try to find it. Commented Feb 28, 2023 at 16:53
• math.stackexchange.com/a/291754/815585 Commented Feb 28, 2023 at 16:55
• @Conrad I have taken up your suggestion to pass to the limit directly in the expansion of the binomial. Indeed, this can be done rigorously but it is not totally straightforward as it requires a little trick. I have meanwhile worked out a detailed proof which is elementary in that it does not exploit any tools which are specific for complex analysis such as analytical continuation nor does it assume the validity of this equation already for real numbers. It is just careful estimations which prove this equation in one go for all complex numbers. You can find all this in my own answer, Commented Feb 28, 2023 at 21:29

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.