How to show that $\sum_{k=0}^n\dfrac{z^k}{k!}\to \mathrm e^z$ without using Cauchy product

Question

I want to show that for $z = x +\mathrm{i}y \in\mathbb{C}$ $$ \sum_{i=0}^n\dfrac{z^i}{i!}\to \left(\sum_{j=0}^n\dfrac{x^j}{j!}\right)\left(\sum_{k=0}^n\dfrac{\mathrm{i}^k}{k!}\cdot y^k\right) $$ (wich tends itself to $\mathrm{e}^z$) as $n\to\infty$ without using the Cauchy product? Is it possible with formal definition of the limit, to show that $$\left\vert \sum_{i=0}^n\dfrac{(x+\mathrm{i}y)^i}{i!} - \left(\sum_{j=0}^n\dfrac{x^j}{j!}\right)\left(\sum_{k=0}^n\dfrac{\mathrm{i}^k}{k!}\cdot y^k\right)\right\vert \to 0 $$ Thanks in advance for help.

Answer 1

Why not just use the "McLaurin series" expansion? It is easy to see that all derivatives of $e^z$ are $e^z$ which are all equal to 1 at x= 0. The McLaurin series expansion of $e^z$ is $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n= \sum_{n=0}^\infty \frac{z^n}{n!}$. Of course, you would still need to prove that this expansion converges to $e^z$ for all z but that is easy using the fact that a power series can be differentiated "term by term". The derivative of $\sum_{n=0}^\infty \frac{z^n}{n!}$ is $\sum_{n=1}^\infty \frac{nz^{n-1}}{n!}= \sum_{n=1}^\infty \frac{z^{n-1}}{(n-1)!}$ (the sum now starts at n= 1 because the 0 term was 1, a constant, and its derivative is 0). Letting m= n-1, that becomes $\sum_{m=0}^\infty \frac{z^m}{m!}$, the same sum as before. That is, $y= \sum_{n=0}^\infty \frac{z^n}{n!}$ satisfies $y'= y$ as well as the initial condition y(0)= 1. And it is well known that that initial value problem has the unique solution $y= e^z$.

Answer 2

You can use the binomial expansion and show $$\left(1 + \frac{z}{N} \right)^N \rightrightarrows \sum_{k=1}^\infty \frac{z^k}{k!} $$ locally uniformly in $\mathbb{C}$ (Actually, for your question i think point-wise convergence in $\mathbb{R}$ is enough).

Then note: $$\sum_{k=1}^\infty \frac{(x+iy)^k}{k!} \leftleftarrows \left(1 + \frac{x+iy}{N}+\frac{ixy}{N^2} \right)^N=\left(1 + \frac{x}{N} \right)^N \cdot \left(1 + \frac{iy}{N} \right)^N \rightrightarrows \sum_{k=1}^\infty \frac{x^k}{k!}\cdot \sum_{k=1}^\infty \frac{(iy)^k}{k!}$$

To show the left limit, note that there is a $O\left(\frac{1}{N^2}\right)$ term (which changes nothing).