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

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

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

$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)$

We can get rid of the double sum as follows:

$ = \sum_{k=0}^{n-1}\left(\prod_{j=0}^k\dfrac{n-j}{1+j}\right)(-z)^{-k} $

But note that here the summands depends on the upper bound of the sum and for this we can't take the limit $n\to \infty$ of the summand term anymore.

There are some related sums, e.g.

expFactor[n_] = (n - (k - j))/(j*f[n]);
invFactor[n_] = -(n - j)/(j + 1);
Sum[Product[expFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify
Sum[Product[invFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify

or written differently

Sum[Product[1 - (k - j)/n, {j, 1, k}] z^k/k!, {k, 0, n}] // Simplify

Sum[Product[-(n - j), {j, 1, k}] (1/(k + 1)) z^k/k!, {k, 0, n}] // Simplify

Wikipedia: Geometric series, Geometric progression

Infinite sum of complex numbers

Finite geometric series