TITLE: Easy explanation of analytic continuation
QUESTION [39 upvotes]: Today, as I was flipping through my copy of Higher Algebra by Barnard and Child, I came across a theorem which said,

The series $$ 1+\frac{1}{2^p} +\frac{1}{3^p}+...$$ diverges for $p\leq 1$ and converges for $p>1$.

But later I found out that the zeta function is defined for all complex values other than 1. Now I know that Riemann analytically continued this function to fit all complex values, but how do I explain, to a layman, that $\zeta(0)=1+1+1+...=-\frac{1}{2}$?
The Wiki articles on these topics go way over my head. I'd appreciate it if someone can explain it to me what analytic continuation actually is, and which functions can be analytically continued?

Edit
If the function diverges for $p\leq1$, how is WolframAlpha able to compute $\zeta(1/5)$? Shouldn't it give out infinity as the answer?

REPLY [5 votes]: Answer to the added paragraph only. I guess at least this part is simple. In the case when the argument of the zeta function is a real $x>1$ the zeta function can be expressed as 
    $$\zeta (x)=\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\frac{1}{1-2^{1-x}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}},\qquad(\ast)$$
where the last series is the Dirichlet eta function. Since for $x>0$ this alternating series converges (because $1/k^x\to 0$), and $1-2^{1-x}\neq 0$ for $x\in]0,1[\cup]1,\infty[$, the function $\zeta (x)$ can be continued analytically to $x\in]0,1[\cup]1,\infty[$ (note that $1$ is excluded as in the first series) as
$$\zeta (x)=\frac{1}{1-2^{1-x}}\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}}.$$
For $x=1/2$, we get $\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx 0.6049$ and
$$\zeta (1/2)=-\left( 1+\sqrt{2}\right) \sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{1/2}}\approx -1.4604.$$
Added: Derivation of $(\ast)$:
$$\begin{eqnarray*}
\zeta (x) &=&\sum_{k=1}^{\infty }\frac{1}{k^{x}}=\sum_{k=1}^{\infty }\frac{%
(-1)^{k-1}}{k^{x}}+2\sum_{k=1}^{\infty }\frac{1}{\left( 2k\right) ^{x}} \\
&=&\sum_{k=1}^{\infty }\frac{(-1)^{k-1}}{k^{x}}+2^{1-x}\zeta (x).
\end{eqnarray*}$$
The relation is as follows.