Why is analytic continuation practical? I was exploring the Riemann Zetta function, and  I observed that $\zeta (s)$ is not normally defined for  $s $ such that $\Re (s) \leq 1$, but analytically continued to the whole complex plane. And the famous Riemann Hypothesis is all about the behavior of the function in the region $0 < \Re(s) < 1$. This means we have accepted the validity of analytic continuation.
Here is my question: why is it valid to assume that a function behaves in the analytically continued domain as it behaves where it is normally defined ? I mean analytic continuation results in counter-intuitive conclusions like "the sum  of all natural numbers is a negative fraction", yet we still use it in our mathematics. Why is that a valid assumption ?
I hope my question is clear. If any of my statements sound absurd or incorrect, I apologize; that's because I just discovered about analytic continuation.
Thanks.
Edit: My question in short is why is $\zeta (-1) = \frac{-1}{12}$ when it should be infinity ?
 A: It doesn't have to be zeta function to realize the
joy of analytic continuation.
You will need to know some background knowledges on power series.
Consider the old geometric series
$$
F(x) = 1 + x + x^{2} + \cdots
$$
as a function of $x$. Just as $\zeta(s)$ sounds nonsense for $s = -1$,
so does, say, $F(2)$.
But, we know that $F(x)$ is written as
$$
F(x) = 1/(1 - x)
$$
for $|x| < 1$.
Now, analytic continuation. The series
$$
1 + x + x^{2} + \cdots
$$
makes sense for $|x| < 1$ (and, differentiable term by term)
and so does the expression 1/(1 - x) for all complex numbers
$x$ except for $x = 1$.
Therefore, F(x) can be uniquely defined (yes, we are
defining) as
F(x) = 1/(1 - x)
for all complex numbers $x \not = 1$.
As you recall that F(x) was nonsense for x = 2 at first,
this sounds a big advance, doesn't it?
Roughly speaking, this depends on the uniqueness of power series
representation; that is, suppose you have two functions
$f(x) = \sum a_{n}x^{n}$ defined on $x \in X$ and
$g(x) = \sum b_{n}x^{n}$ on $x \in Y$, $X \subset Y$,
that are equal over some intersection of $X$ and $Y$.
Well, power series are unique, so f is g for the larger domain $Y$
as well.
