Calculate the power series centered in $z_0=1$ I did this, I don't know if is correct
$$
\begin{gathered}
h(z)=\left(\frac{z}{z+1}\right)^{2} \\
u=z-1,z_0=1 \\
h(z)=\left(\frac{z}{z+1}\right)^{2}=z^{2} \frac{1}{(z+1)^{2}}=z^{2} \frac{1}{(1-(-z))^{2}} \rightarrow \text { geometric series } \\
\sum_{n=0}^{\infty} u^{n}=\frac{1}{1-u}, \quad|z|<1, \\
\\\frac{d}{d z} \frac{1}{1-u}=\frac{1}{(1-u)^{2}}
\\\left(\frac{z}{z+1}\right)^{2}=z^{2} \frac{d}{d u} \sum_{n=0}^{\infty} u^{n}=z^{2} \sum_{n=0}^{\infty} n u^{n-1}=z^{2} \sum_{n=0}^{\infty} n(z-1)^{n-1}=\sum_{n=0}^{\infty} n(z-1)^{n+1} \\
\left(\frac{z}{z+1}\right)^{2}=\sum_{n=0}^{\infty} n(z-1)^{n+1},|z-1|<1 \rightarrow|z|<2 \\
\\
\end{gathered}
$$
I need to use a geometric serie to Calculate the power series
I have a problem, the problem is in $z_0=1$  I don't know if I use it right
 A: Comment on your approach:

It is not very clear how each step performs, in addition the changes of variables are strange in the part that tries to adjust to the convergence radius. We can use only geometric series if we first make use of partial fractions and then study each series representation for each function obtained and we adjust the radius of convergence.

First use partial fraction
$$\left(\frac{z}{z+1}\right)^{2}=1-\frac{2}{z+1}+\frac{1}{(z+1)^{2}}$$
Now,

*

*The series expansion of a constant around every point is the same constant, in this case
$$\boxed{1=1,\quad \forall z\in \mathbb{C}}$$

*By Geometric Series we know that,

$$\sum_{n=0}^{+\infty}az^{n}=\frac{a}{z-1},\quad a\in \mathbb{C},\quad |z|<1$$
Now notice that
$$\frac{{\rm d}}{{\rm d}z}\left(\frac{-1}{z+1}\right)=\frac{1}{(1+z)^{2}}$$
and also,
\begin{align*}
\frac{-1}{z+1}&=\left(-\frac{1}{2}\right)\left(\frac{1}{1-\left(\frac{-z+1}{2}\right)}\right),\\ &=\left(-\frac{1}{2}\right)\sum_{n=0}^{+\infty}\left(\frac{-z+1}{2}\right)^{n},\\ &=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}(z-1)^{n},\quad |z-1|<2
\end{align*}
Then,
\begin{align*}
\frac{1}{(1+z)^{2}}&=\frac{{\rm d}}{{\rm d}z}\left(-\frac{1}{z+1}\right),\\&=\frac{{\rm d}}{{\rm d}z}\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}(z-1)^{n},\\&=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}n(z-1)^{n-1},\\
&=\sum_{n=1}^{+\infty}\left(-\frac{1}{2}\right)^{n+1}n(z-1)^{n-1},\\
&\overset{k=n-1}{=}\sum_{k=0}^{+\infty}\left(-\frac{1}{2}\right)^{k+2}(k+1)(z-1)^{k},\\
&\overset{k=n}{=}\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n}
\end{align*}
where the differentiation is allowed because a power series can be differentiated term by term in any region that lies entirely inside its circle of convergence.
Therefore,
$$\boxed{\frac{1}{(1+z)^{2}}=\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\quad |z-1|<2}$$

*

*Again by Geometric Series we have
\begin{align*}
-\frac{2}{z+1}&=2\left(-\frac{1}{2}\right)\left(\frac{1}{1-\left(\frac{-z+1}{2}\right)}\right),\\&=-\sum_{n=0}^{+\infty}\left(\frac{-z+1}{2}\right)^{n},\\ &=\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n},\quad |z-1|<2
\end{align*}
Therefore,
$$\boxed{-\frac{2}{z+1}=\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n},\quad |z-1|<2}$$
If $|z-1|<2$, we have
\begin{align*}
\left(\frac{z}{z+1}\right)^{2}&=1-\frac{2}{z+1}+\frac{1}{(z+1)^{2}},\\&=1+\sum_{n=0}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n}+\sum_{n=0}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\\&=1-1+\sum_{n=1}^{+\infty}-\left(-\frac{1}{2}\right)^{n}(z-1)^{n}+\frac{1}{4}+\sum_{n=1}^{+\infty}\left(-\frac{1}{2}\right)^{n+2}(n+1)(z-1)^{n},\\
&=\frac{1}{4}+\sum_{n=1}^{+\infty}\left[-\left(-\frac{1}{2}\right)^{n}+\left(-\frac{1}{2}\right)^{n+2}(n+1) \right](z-1)^{n},\\
&=\frac{1}{4}+\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{2^{n+2}}(n-3)(z-1)^{n}
\end{align*}
Therefore the series expansion at $z=1$ is given by,
$$\boxed{\left(\frac{z}{z+1}\right)^{2}=\frac{1}{4}+\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{2^{n+2}}(n-3)(z-1)^{n},\quad |z-1|<2}$$
Expanding we get
$$\left(\frac{z}{z+1}\right)^{2}=\frac{1}{4}+\frac{1}{2^{2}}(z-1)-\frac{1}{2^{4}}(z-1)^{2}+\frac{1}{2^{6}}(z-1)^{4}+\mathcal{O}(z-1)^{5}$$
