Power series of $\frac{1}{(1-z)^2}$ I want to show, that the following is true for every $z\in C$ with $|z|<1$:
$$\frac{1}{(1-z)^2} =\sum_{k=1}^\infty kz^{k-1}$$
I think there is a way with the Cauchy-Product
 A: Since the series expansion of $\frac{1}{1-z}$
is
$\frac{1}{1-z} =\sum_{k=0}^\infty z^{k}$
for every $z\in C$ with $|z|<1$ differentiating both sides with respect to $z$ you will get
$\frac{1}{(1-z)^2} =\sum_{k=1}^\infty kz^{k-1}$ for $|z|<1$.
A: Since
$$
\frac{1}{1-z}=\sum_{k=0}^{\infty}z^k
$$
the Cauchy product tells you that
$$
\frac{1}{(1-z)^2}=
\sum_{k=0}^{\infty}
\biggl(\,\sum_{l=0}^k1\biggr)z^k=
\sum_{k=0}^\infty (k+1)z^k=
\sum_{k=1}^\infty kz^{k-1}
$$
A: It is apparent that the series is absolutely convergent for $\lvert z\rvert<1$. In which case, notice that $g(z)=\sum_{k=0}^\infty (k+1)z^k=\sum_{k=0}^\infty z^k+\sum_{k=1}^\infty kz^k=\frac1{1-z}+zg(z)$
A: You're absolutely right that this can be proved with the Cauchy product. The derivative approach is nice, but involves a lot of machinery (and if you can't prove the theorems necessary then it's best to look for a different way). Informally, we can write
\begin{align}
\frac{1}{(1-z)^{2}} &= \frac{1}{1-z} \frac{1}{1-z} \\
&= (1+z+z^{2}+z^{3}+\ldots)(1+z+z^{2}+z^{3}+\ldots) \\
&= 1 + (z+z) + (z^{2}+z\cdot z + z^{2}) + \ldots\\
&= 1+2z+3z^{2} + \ldots
\end{align}
To see that there are exactly $k+1$ copies of each $z^{k}$ in the resulting sum, match up the terms in the individual products as follows: $z^{0}$ in product one pairs up with $z^{k}$ in product 2, $z^{1}$ pairs up with $z^{k-1}$, and so on... $z^{k}$ in product 1 pairs up with $z^{0}$ in product 2. This is all the ways of making $z^{k}$, and there are $k+1$ such combinations. So the coefficient of $z^{k}$ is $k+1$, as required. 
In symbols, we have 
$$\frac{1}{1-z} = \sum_{n=0}^{\infty}a_{n} z^{n}$$
The Cauchy formula gives
$$\left(\sum_{i=0}^{\infty} a_{i}z^{i}\right)\left(\sum_{j=0}^{\infty} b_{j}z^{j}\right) = \sum_{n=0}^{\infty} c_{n}z^{n}$$
where $$c_{n} = \sum_{k=0}^{n} a_{k}b_{n-k}$$
In our case, $a_{k} = b_{k} = 1$ for every $k$, so the sum is obviously $c_{n} = n+1$, as required.
