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 $$I\_n=\int\_1^e\frac{(\ln x)^n}{x^2}dx.$$ Using integration by parts we have $$I\_{n+1}=-\frac{1}{e}+(n+1)I\_n.$$
Using induction we can prove that
$$\frac{I\_n}{n!}=1-\frac{1}{e}\sum\_{k=0}^n \frac{1}{k!}$$
since $(I\_n)\_n$ is bounded, we let $n\to \infty$ we get
$$\sum\_{k=0}^\infty \frac{1}{k!}=e.$$