How do I prove that $e^x = \sum_{i \geq 0} x^i/i!$ I want to show this using epsilon-N argument
What $N$ should I choose so that if $m > N \implies$ $\left | e^x - \sum_{i=0}^{m} x^i/i! \right | < \epsilon $.
My domain is $\forall x \in \mathbb{R}$.
My goal is to show that this "peculiar" (Taylor) sum converges to $e^x$, not trying to show that this sum converges.
 A: I propose the following:
1) Prove the power series converges (absolutely, uniformly on compact sets and whatever) for all $\;x\in\Bbb R\;$ and thus defines a (continuous and derivable) real function $\,e(x)\;$.
2) Differentiate now elementwise the series:
$$e'(x)=\sum_{n=1}^\infty\frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty\frac{x^{n-1}}{(n-1)!}=\sum_{n=0}^\infty\frac{x^n}{n!}=e(x)$$
(3) Solving the above differential equation you get $\;e(x)=ke^x\;,\;k\in\Bbb R\;$ , and since $\,e(0)=1\implies k=1\;$ and we're done .
The above is, imo, very nice, but in some sense it sucks since we rely on differential equations which, at this stage, may not be the most natural way to follow... but OTOH it is a pretty simple diff. eq. and, after all, even at high school level sometimes we define "The system E(X)", which is basically the above (and other stuff, too.)
A: The way this is often done follows closely DonAntonio's answer, but we can avoid the use of the theory of differential equations as follows. Let $F(x)$ be the sum of this series. Then it follows that $F'(x)=F(x)$ for all $x$, and $F(0)=1$. Now, assume that $G(x)$ is another function that satisfies these assumptions, e.g.
$G(x)=e^x$, if you believe that $e^x$ can be defined and differentiated by some other means :-)
Then let's differentiate:
$$
D\left(\frac{F(x)}{G(x)}\right)=\frac{F'(x)G(x)-G'(x)F(x)}{G(x)^2}
=\frac{F(x)G(x)-G(x)F(x)}{G(x)^2}=0
$$
for all $x$. By mean value theorem (our sub for the solutions of that DE!) we know that $F(x)/G(x)$ is a constant. Setting $x=0$ we find that the constant is equal to $1$. Therefore $F(x)=G(x)$ for all $x$, in other words the above properties determine the function $F(x)=e^x$ uniquely.
