# What is the sum $\sum_{k=0}^{\infty} kz^{-k}$?

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

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

would be very appreciated.

• 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? – Thomas Andrews Apr 8 at 15:35
• Is your question : is there a closed formula for this series ? – Jean Marie Apr 8 at 15:37
• Yeah, I am tired. Thank you, I solved my problem within 2sec when I saw your suggestion, @ThomasAndrews – OSCAR Apr 8 at 15:38

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 $$|z|>1.$$

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,

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