I'm struggling with getting the power series/Taylor series expansion of $f(z) = \frac{1}{(1+z)^2}$ around $z_0 = 1$.
Usually, I would do a partial fraction decomposition, and then do some re-arranging of the expression to get the geometric series expansion, but can't seem to be able to do that here.
So far I have $$f(z) = \frac{1}{4} \left( \frac{1}{1 - \left( - \frac{z-1}{2} \right)} \right)^2 = \frac{1}{4} \left( \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^k} (z-1)^k \right)^2$$
and I don't know what to do about the square around the sum.
Observe that $f(z) = 1/(1+z)^2=1/((z-1)+2)^2=\frac{1}{4((z-1)/2+1)^2}$.
Now let $(z-1)/2 := u$, and suddenly we have the Taylor expansion for $f(u) = 1/(4 \cdot (u+1)^2)$ but around $u = 0$.
Realize that $f(u) = -1/4 \cdot \frac{d}{du}\left(\frac{1}{1+u}\right) = -1/4 \cdot \frac{d}{du}\left(\frac{1}{1-(-u)}\right)$. We already know that
$$h(u) = \frac{1}{1-u} = \sum_{k\geq 0} u^k$$ for $|u|<1$. So $f(u) = -1/4 \cdot h(-u) = -1/4 \cdot \sum_{k\geq 0}(-1)^k u^k$. Thus, we want $f(u) = -1/4 \cdot h'(-u)$.
Hence $f(u) = -1/4 \cdot \frac{d}{dz}\sum_{k\geq 0}(-1)^k u^k$. Since we have two limits essentially, one from the derivative, and one of the infinite sum, we have to switch the order of the limits. This can be done if we can show that the sum is uniformly convergent for $|u| < 1$ or equivalently $|z-1|<2$.
In this case, it turns out to be the true. I won't go into showing that, since there's lots of proofs for this simple case out there. Either way, substituting back for $u = (z-1)/2$, we get $f(z) = -2^{-2} \cdot \sum_{k\geq 0}(-1)^k k ((z-1)/2)^{k-1} = \sum_{k\geq 1}(-1)^{k-1} 2^{-1-k}k (z-1)^{k-1} = 1/4-1/4\cdot (z-1)+3/16\cdot(z-1)^2-...$