EDIT: $a_0 = $ the first term in a series, $r=$ the common ratio in a gemoetric series.
I've done many analytic continuations and keep coming to the same conclusion; the analytically continued sum always evaluates to the value of $\frac{a_0}{1-r}$, as long as $r$ does not equal $1$ at the point of evaluation, causing a pole. For instance, in the sum:
$1+2+4+8+16+32...\ \rightarrow\ 1-x+x^2-x^3+x^4-x^5...\ : x=-2$
$a_0 = 1,\ r=-x,\ \frac{a_0}{1-r}=\frac{1}{x+1}$
At $x = -1$, we get a pole, so we cannot evaluate $1+1+1+1+1+1...$, but we can evaluate at $-2$. I did all of the work to analytically continue our convergence radius to include $-2$(skipping over the pole at $-1$), and I got $-1$. $-1$ is also the solution to $\frac{a_0}{1-r}$ where x is $-2$. So basically, for this example, I just showed that you can completely disregard the rule of $|r|<1$ that comes with $\frac{a_0}{1-r}$. I also tested this for Grandi's series, which I used the same formula, $\frac{a_0}{1-r}=\frac{1}{x+1}$, evaluates at $x=1$. I analytically continue and get $1/2$. Once again, we get that we just disregard $|r|<1$ "shorthand" analytic continuation. Is this true? Do I actually have to do all of the work with Taylor Series, or can I just evaluate $\frac{a_0}{1-r}$?