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Any hint clarifying the problem as stated in the title, i.e

what is $\sum_{k=0}^{\infty} kz^{-k}$?

would be very appreciated.

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    $\begingroup$ Let $w=\frac{1}{z}$ then you get $\sum_{k=0}^{\infty} kw^k.$ This is a more well-known form (but still might not be enough for your answer.) What ha e you tried? $\endgroup$ – Thomas Andrews Apr 8 at 15:35
  • $\begingroup$ Is your question : is there a closed formula for this series ? $\endgroup$ – Jean Marie Apr 8 at 15:37
  • $\begingroup$ Yeah, I am tired. Thank you, I solved my problem within 2sec when I saw your suggestion, @ThomasAndrews $\endgroup$ – OSCAR Apr 8 at 15:38
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Hint: Letting $w=\frac{1}{z}$ this is the same as: $$\sum_{k=0}^{\infty} kw^k.$$

This is a slightly more well-known series.

This will converge when $|w|<1$ and hence when $|z|>1.$

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Hint:

Consider the entire function of $z^{-1}$

$$f(z^{-1})=\sum_{k=0}^\infty(z^{-1})^k,$$

which converges to $$\frac1{1-z^{-1}}$$ for all $|z|>1$.

Now by termwise differentiation,

$$f'(z^{-1})=\sum_{k=0}^\infty k(z^{-1})^{k-1}=z\sum_{k=0}^\infty k(z^{-1})^k.$$

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