Sum of geometric progression $z^{-m}$?

Question

We know that the sum of the geometric progression of $z^{-n}$ where $n$ starts at zero and goes to infinity is $1/(1-z)$, but is there a way to show the sum when $n$ starts at $m$? I.e. what is

$$\sum_{n=m}^\infty \frac{1}{z^n}$$

Answer 1

$$ \sum_{k=m}^\infty z^{-k} = \sum_{k=0}^\infty z^{-m - k} = z^{-m}\sum_{k=0}^\infty z^{-k} $$

Answer 2

$\frac{1}{1-z}$ is the sum of $z^k$ (when $|z|<1$), NOT $z^{-k}$ (in this case the sum would be $\frac{z}{z-1}$ and $|z|$ would be greater than $1$). However if $|z|<1$ we have that

$$ \sum_{k=m}^{+\infty}z^{k}= \sum_{k=0}^{+\infty}z^{k}-\sum_{k=0}^{m-1}z^{k}=\frac{1}{1-z}-\frac{1-z^m}{1-z} =\frac{z^m}{1-z} $$

If you mean $z^{-k}$ changes only a few: as already said $|z|>1$ and you'd have just to turn $z$ into $\frac{1}{z}$ in the above formula: $$ \sum_{k=m}^{+\infty}z^{-k}=\frac{\frac1{z^m}}{1-\frac1{z}}=\frac{1}{z^m-z^{m-1}} $$