How to show that $\sum_{k=0}^{\infty}{(k+1)z^k}=\frac{1}{(1-z)^2}$ for $z\in \mathbb{C}:|z|<1$

Question

How to show that $\sum_{k=0}^{\infty}{(k+1)z^k}=\frac{1}{(1-z)^2}$ for $z\in \mathbb{C}:|z|<1$

I believe that I have to use the cauchy product?

But how do transform the expression to be a product in the first place?

Answer 1

You mention Cauchy product, but don't say what you try. Let $(1-z)^{-1} = \sum a_{n} z^{n} = \sum b_{n} z^{n}$.

Then, in the usual notation, \begin{align*} (1-z)^{-2} &= \left(\sum_{n=0}^{\infty}a_{n}z^{n}\right)\left(\sum_{n=0}^{\infty}b_{n}z^{n}\right) \\ & = \sum_{n=0}^{\infty} c_{n}z^{n} \\ \text{where}\quad c_{n}&= \sum_{i=0}^{n}a_{i}b_{n-i} \\ &= \sum_{i=0}^{n}1 \\ &= n+1 \end{align*}

exactly as required.

Answer 2

$$f(z)=\frac1{1-z}=\sum_{k=0}^\infty z^k\,,\,\,|z|<1\implies f'(z)=\frac1{(1-z)^2}=\left(\sum_{k=0}^\infty z^k\right)'\,,\,|z|<1\ldots$$