Power series of a function around a point ≠ 0 Using the power series $ \sum_{k \geq 0}z^k = \frac{1}{1-z} $ for $ |z|<1 $, where it's centered around $0$. How would the series look like if someone wanted to let, for example $ z_0 = \frac{1}{2}$?
Also, how would one compute the power series of $ \frac{1}{1+z^2} $ around $z_0 = 1$?
I know that, centered around 0 we can manipulate the geometric series to obtain $ \frac{1}{1+z^2} = \sum_{k \geq 0} (-1)^k z^{2k} $. However, I want to write it on the form $ \sum_{k \geq 0} c_k (z-1)^k $.

Usually, when I want a series about a point $z_0 \ne 0$, I change variables and write $w = z-z_0, z = w+z_0$ and do the series in powers of $w$.
Example
$\frac{1}{1+z^2}$ centered at $z=1$.  Write $w=z-1, z=w+1$, so
$$
\frac{1}{1+z^2} = \frac{1}{1+(w+1)^2} = \frac{1}{2+2w+w^2}
$$
For this I could use partial fractions
$$
\frac{1}{2+2w+w^2} = \frac{i/2}{w+(1+i)} - \frac{i/2}{w+(1-i)}
$$
and each of these is a geometric series.  It turns out (as expected) that
the imaginary parts all cancel.  We can see a pattern:
$$
\frac{1}{2} -\frac{1}{2}(z-1) 
+\frac{1}{4}(z-1)^2 
-\frac{1}{8}(z-1)^4 +\frac{1}{8}(z-1)^5 
-\frac{1}{16}(z-1)^6 
+\frac{1}{32}(z-1)^8 -\frac{1}{32}(z-1)^9 
+\frac{1}{64}(z-1)^{10} \dots
$$
