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

$$ e = \sum\_{n=0}^\infty \frac{1}{n!}, $$
Since you are calling **for intuition, a term by term version**. Intuitively, the inverse factorial coefficient $\frac{1}{n!}$ is the most natural, as it yields "**some kind of invariance to differentiation**", since:

$$\left(\frac{x^n}{n!}\right)' = \left(n\frac{x^{n-1}}{n!}\right) = \frac{x^{n-1}}{(n-1)!} = \frac{x^{m}}{m!}$$
with $m= n-1$.

You may know that, and start with, a fundamental property of the natural exponential: it is equal to its derivative. 
So, suppose that  

$$ f(x) = \sum\_{n=0}^\infty a\_n{x^n}\,,$$
is equal to its derivative. Then formally (I am skipping issues on convergence), 
$$ f'(x) = \sum\_{n=1}^\infty n a\_n{x^{n-1}}\,,$$
thus by reindexing:
$$ f'(x) = \sum\_{n=0}^\infty (n+1) a\_{n+1}{x^{n}}\,,$$
then, one should have, term by term:
$$ a\_n= (n+1) a\_{n+1}\,,$$
hence
$$ a\_{n+1} = \frac{a\_n}{n+1} =\frac{a\_0}{(n+1)!}\,.$$
Since the exponential is the reciprocal to the $\log$, you require that $f(0)=1$, hence $a\_0=1$. So naturally, 

$$ f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}=e^x\,,$$
and 
$$ f(1) = \sum\_{n=0}^\infty \frac{1}{n!}=e\,.$$