real analysis - Infinite series and the Riemann zeta function

I have two questions concerning infinite series in the context of the Riemann zeta function.

1. Given the properties of infinite series, why can't we regroup the terms in $\zeta(0)$ in such a way as to give $\zeta(-1)$? i.e.

$$\zeta(0)=\sum_{n=0}^\infty \frac{1}{n^0}=\sum_{n=0}^\infty 1=1+1+1+\ldots=(1)+(1+1)+(1+1+1)+\ldots=1+2+3+\ldots=\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{1^{-3}}+\ldots=\sum_{n=0}^\infty n=\sum_{n=0}^\infty \frac{1}{n^{-1}}=\zeta(-1)$$

1. This one might be a lot simpler to answer: why can we assign a value to $\zeta(-1)=\sum_{n=0}^\infty \frac{1}{n^{-1}}$ when the infinite series on the RHS is clearly divergent, i.e. its $n^{th}$ term is always bigger than its $(n-1)^{th}$ term?

Answer

In short: in a non-absolutely convergent series you can't do things like reorder and group terms because you may get a different answer. In fact, you can reorder the terms in in the sum $1/1-1/2+1/3-1/4+...$ (which in this case does converge, but not absolutely) to give you any real result you like!!!
http://en.wikipedia.org/wiki/Absolute_convergence

$\zeta(-1)$ is something quite different. For numbers with $Re(s) \leq 1$ we don't define $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$, but instead as the function which "smoothly" extends this sum which is well-defined on $Re(s) > 1$ to those numbers with $Re(s) \leq 1$. It is, as was mentioned, a meromorphic continuation.