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

$$ e = \sum\_{n=0}^\infty \frac{1}{n!}, $$
By definition   

$$\frac{d}{dx}e^x = e^x$$
therefore
$$\int\_0^xe^t\*1dt=e^x-1$$
we can use integration by parts which state that:
$$\int{f(x)g(x)dx}=f(x)\int g(x) - f^`(x)\int\int g(x)+...$$
let f(x)= e^x, g(x) = 1
$$e^x\int1-e^x\int\int1+...=e^x-1$$
we simplify to get:
$$\frac{1}{e^x}=1-\int1+\int\int1-\int\int\int1-...$$
we know that $\int\int..n..\int1dx=\frac{x^n}{n!}$ (basic integration rule)   

therefore
$$\frac{1}{e^x}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-...$$

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let substitute x = -n, we know that (-x)^n = -x^n when n is odd
$$e^n = 1+\frac{x^1}{1!}+\frac{x^2}{2!}+..$$
in fact, if we are to substitute x = ln(f(n)), n > 0 we get a nice series:
$$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$