Question: A finite Hausdorff topological group, has discrete topology and every discrete group is totally disconnected. I look for an example of a non-Abelian, finite, connected non-Hausdorff group . I think every non-Abelian, finite, connected groups must be discrete.

Answer: 
These two familiar sums are the Taylor series for $e^x$ about $0$.
To get $e$ itself, you evaluate this series at $x=1.$

**Derivation:** The $n$th term of the [Taylor series](http://mathworld.wolfram.com/TaylorSeries.html) of a function $f$ about $a$ is

$$ \frac{f^{(n)}(a)}{n!} (x-a)^n.$$

But if $f(x) \triangleq e^x$, then $f'(x) = e^x$, and by an inductive argument,
$f^{(n)}(x) = e^x$ for every positive integer $n.$
Taking the series about $a = 0,$ the $n$th term is

$$ \frac{f^{(n)}(a)}{n!} (x-a)^n = \frac{e^a}{n!} (x-a)^n
 = \frac{e^0}{n!} (x-0)^n = \frac{x^n}{n!}.$$

That is, the Taylor series of $e^x$ as a function of $x$ about $0$ is

$$ e^x = \sum\_{n=0}^\infty \frac{x^n}{n!}, $$

and by setting $x=1$ we get

$$ e = \sum\_{n=0}^\infty \frac{1}{n!}, $$
Let $x \in \mathbb{R}^+$.

$f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ :

$$\left|f(x)-\sum\_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I\_{n+1} \text{ where } I\_{n+1}=sup\_{[0,x]}|f^{n+1}|$$

And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I\_{n+1}=\exp(x)$

But $\displaystyle\lim \_{n\to \infty }\left(\frac{|x|^{n+1}}{(n+1)!}\exp(x)\right)=\:0$

So $\displaystyle\lim \_{n\to \infty }\left(\sum\_{k=0}^n\frac{x^{k}}{k!}\right)=\exp(x)$

Just take $x = 1$ for $\mathbb{e}$