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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?

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

DonAntonio
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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.

preferred_anon
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  • Does the part underneath $\text{where} \ldots$ also belong to $\sum_{n=0}^{\infty} c_{n}z^{n}$, so that $\sum_{n=0}^{\infty} c_{n}z^{n}=n+1$ where $c_{n}= \sum_{i=0}^{n}a_{i}b_{n-i}$ – Analysis Jun 16 '19 at 12:45
  • Nope! Each time I write a new LHS all the RHSs until the next LHS belong to it – preferred_anon Jun 16 '19 at 12:54
  • (also, $n$ is of course not a free variable in the first few lines, so it can't be free in any equal lines) – preferred_anon Jun 16 '19 at 12:54