$Q_\infty(z)=\dfrac{1}{1-z}$

This can also be written as

$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^k = 1+\dfrac{1}{z}$

and

$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^k = z$

or, for $z>0$ and $X<1+z$ resp. $X<z/(z-1)$

$\sum_{k=0}^\infty\left(\dfrac{1}{1+z}\right)^kX^k = 1+\dfrac{1}{z}+(X-1)(z-1) \,z\dfrac{1}{z+X(1-z)}$

and

$\sum_{k=0}^\infty\left(1-\dfrac{1}{z}\right)^kX^k = z+(X-1)\left(1+\dfrac{1}{z}\right)\dfrac{1}{z+(1-X)}$

For a function $f$, the q-integral from $0$ to $1$ (“$z$-integral” if we stick to our notation above) is defined as

$\sum_{k=0}^\infty f(z^k)\,z^k=\dfrac{1}{1-z}\cdot\int_0^1 f(s)\,{\mathrm d}_zs$

$z = \sum_{k=0}^\infty\left(z^{-1}(z-1)\right)^k = \sum_{k=0}^\infty\left(1-z^{-1}\right)^k = \sum_{k=0}^\infty \sum_{m=0}^k {k \choose m}(-z)^{-m} $

In fact

$ \sum_{k=0}^{n-1} \sum_{m=0}^k {k \choose m}(-z)^{-m} = z\left(1-\left(\dfrac{z-1}{z}\right)^n\right)$

See also Niemand sequences.

Sum[Binomial[-s, k] x^k, {k, 0, \[Infinity]}]
Series[(1 - x)^-s, {x, 0, 4}]
Binomial[-s, 3];
% - (-s)!/(3! ((-s) - 3)!) // FullSimplify
%% + 1/6 (2 s + 3 s^2 + s^3) // FullSimplify

Wikipedia: Geometric series, Geometric progression

Infinite sum of complex numbers

Finite geometric series