Riemann Zeta Function's Analytic Continuation I am doing some independent learning on the Riemann Zeta function just out of curiosity.  I have searched the site, and I know that there are plenty of questions on the Riemann Zeta function.  However, this particular one has not been answered for me nor have I been able to find it easily on google.  I teach high school math and adjunct at a local community college in lower level classes such as Trig and College Algebra.  I say that so that you would keep in mind the level of mathematics that I have been exposed to and that you will word your answers accordingly (at an elementary level with little sophistication).  I understand that it is claimed that the zeta function is only valid for Re(s)>1 which is why negative arguments require analytic continuation.  But why is the function as it is written not defined for Re(s)&lt;1?  Is this because the series diverges when Re(s)&lt;1?  And if so, why may we claim that the function is undefined for those values just because it is divergent for those values?  It will really take some convincing for me to believe that we can change a function just because it is divergent at some values.
Let's begin by presenting a simple example of representing a function with a series.  Let $f(z)=\frac{1}{1-z}$ for $z\ne 1$.  Recall that $f(z)$ can be represented by the geometric series
$$f(z)=\sum_{n=0}^\infty z^n \tag 1$$
for $|z|&lt;1$.  So, although $f(z)$ exists for all $z\ne 1$, its representation as given in $(1)$ is valid only when $|z|&lt;1$.
We can represent also represent $f(z)$ by the series
$$f(z)=-\sum_{n=1}^\infty \left(\frac{1}{z}\right)^n \tag 2$$
for $|z|&gt;1$.  So, we have two representations for the same function that are valid in distinct regions of the complex $z$-plane.

Now, suppose that we represent the function denoted $\zeta(s)$ by the series
$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$
for $\text{Re}(s)&gt;1$.  We can easily extend the definition by writing
$$\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^s}}_{\zeta(s)}=\underbrace{2\sum_{n=1}^{\infty}\frac{1}{(2n)^s}}_{2^{1-s}\zeta(s)} -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}\tag 3$$
for $\text{Re}(s)&gt;1$.  Upon rearranging $(3)$ we find
$$\zeta(s)=(1-2^{1-s})^{-1}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\tag 4$$
But notice that the series on the right-hand side of $(4)$ converges for $\text{Re}(s)&gt;0$.  So, we have just developed another representation for $\zeta(s)$ that is valid in a larger region of the complex $s$-plane.
There are other series representations of the Riemann Zeta function, such as its Laurent series,
$$\zeta(s)=\frac1{s-1}+\sum_{n=1}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$$
which converges for all $s\ne 1$.
And there are integral representation of $\zeta(s)$ such as
$$\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx$$
which converges for $\text{Re}(s)&gt;1$ and (see 25.5.11 this reference)
$$\zeta(s)=\frac12 +\frac{1}{s-1}-2 \int_0^\infty \frac{\sin(s\arctan(x))}{(1+x^2)^{s/2}(e^{2\pi x}+1)}\,dx$$
which is valid for all $s\ne 1$.
When the real part is more than one, the total length of the pieces $|n^{-k}|$ is finite.  So there is a well-defined sum.
When the real part is between zero and one, the pieces shrink to zero, but the total length of the pieces is infinite.  As a result, you can rearrange the terms and form any sum you like.
For example, 
$$1-\frac12+\frac13-\frac14+...=\ln 2\\
(1+\frac13-\frac12)+(\frac15+\frac17-\frac14)+(\frac19+\frac1{11}-\frac16)+...&gt;(1+\frac13-\frac12)=5/6$$

When $k&lt;0$, so the pieces are $n^{-k}$ with a positive exponent, the pieces get bigger and bigger, so as you say can't settle to anything.  

A real function can change its form as you cross a point.  There is a function that is zero if $x\leq0$, but $\exp(-1/x^2)$ when $x&gt;0$  This function is quite smooth at zero.  

A complex function can't change suddenly.  When a complex function is differentiable, there is only one way to extend it.  That is second-year calculus.  

We have $f(x)=\sum n^{-k}$ defined for $\Re(k)&gt;1$ and $g(x)$ defined everywhere.  $f(x)=g(x)$ when $\Re(k)&gt;1$, and there is only one way to extend it.  So we call the sum $g(x)$, even though the sum doesn't converge.