Power series of $\frac{1}{(1-z)^2}$

Question

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

Answer 1

Since $$ \frac{1}{1-z}=\sum_{k=0}^{\infty}z^k $$ the Cauchy product tells you that $$ \frac{1}{(1-z)^2}= \sum_{k=0}^{\infty} \biggl(\,\sum_{l=0}^k1\biggr)z^k= \sum_{k=0}^\infty (k+1)z^k= \sum_{k=1}^\infty kz^{k-1} $$

Answer 2

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 $|z|<1$.

Answer 3

It is apparent that the series is absolutely convergent for $\lvert z\rvert<1$. In which case, notice that $g(z)=\sum_{k=0}^\infty (k+1)z^k=\sum_{k=0}^\infty z^k+\sum_{k=1}^\infty kz^k=\frac1{1-z}+zg(z)$

Answer 4

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.