Analytic continuation of a function Let 
$$f(z) = A_0 + A_1(z-a) + A_2(z-a)^2 + \cdots$$
converge in some disk $|z - a| < r$. Following Weyl, we magically re-arrange this power series at point $b$ in this disk and the power series should converge in a disk of radius $r - |b-a|$. If it converges outside this disk, we have an analytically continuation of our function outside it's original disk of convergence.
This is all definitions though, how would you actually do this with a few explicit examples (preferably ones that capture the spirit of what's going on here, e.g. sums, integrals etc... and not just the standard geometric series example - is that the only one that exists?) 
thanks!
 A: The reason that you don't see more examples is that the coefficients of the rearranged series cannot be computed recursively in a finitary way, as with the product or the composition of two series. In general each  coefficient $B_k$ of the rearranged series $\sum_{k=0}^\infty B_k(z-b)^k$ is the sum of an infinite series involving the $A_k$: One has
$$B_n={f^{(n)}(b)\over n!}={1\over n!}\sum_{k=0}^\infty A_k\> k(k-1)\cdots(k-n+1)(b-a)^{k-n}=\sum_{k=n}^\infty{k\choose n}\>A_k\>(b-a)^{k-n}\ .$$
Therefore all examples you can find are "special", in the sense that there is an overriding principle at work (a functional equation for $f$, a differential equation, etc.) that gives you the $B_k$ by some other means.
Here is one such example:
$$f(z):={1\over 1+z^2}=\sum_{j=0}^\infty (-1)^j\>z^{2j}\qquad\bigl(|z|<1\bigr)\ .$$
Expanding this function at the point $b:=1$ instead of $a:=0$ gives
$$f(z)={1\over2}-{1\over2}(z-1)+{1\over4}(z-1)^2-{1\over8}(z-1)^4+{1\over8}(z-1)^5-{1\over16}(z-1)^6+{1\over32}(z-1)^8-{1\over32}(z-1)^9+{1\over64}(z-1)^{10}-\ldots\ ,$$
where it is not difficult to detect the law of coefficients. (Note that there are no terms of degree $4j-1$.) The rearranged series has radius of convergence $\sqrt{2}$, and converges, e.g., for $z:=2$, where the original series is no longer valid.
A: Let me describe a procedure for you to be able to answer your own question:
First, use the infinite geometric sum to expand $\frac{1}{1-z}$ and recall that this sum converges only for $z$ within the unit disk. Next, simply take the derivative on both sides. By analytic continuation, your derivative can be defined for all complex $z$ except for 1. Use your result to determine the "value" of $$1-2+3-4+5-6+\cdots $$ I write "value" in quotations since this sum diverges, so the value only applies under the analytic continuation.
