What does the following sum converge to? is there a closed-form formula?

Question

Is there a closed-form term for the following sum:

$$\sum_{i=z}^{\infty}y^{i}\frac{1-(\frac{x}{y})^{(i+1)}}{1-(\frac{x}{y})},$$

where $x<y<1$ and $z$ is integer grater than $0$.

Answer 1

Yes, just use the formula $$\sum_{n=0}^\infty \alpha^n=\frac{1}{1-\alpha}$$ for $|\alpha|<1$. We have that $$\begin{align} \sum_{i=z}^\infty y^i \frac{1-(x/y)^{i+1}}{1-x/y} &=\frac{1}{1-x/y}\sum_{i=z}^\infty y^i-\frac{x}{y-x}\sum_{i=z}^\infty x^i\\ &=\frac{1}{1-x/y}\frac{y^z}{1-y}-\frac{x}{y-x}\frac{x^z}{1-x}\\ &=\frac{y^{z+1}}{(1-y)(y-x)}-\frac{x^{z+1}}{(1-x)(y-x)}\\ &=\frac{1}{y-x}\bigg(\frac{y^{z+1}}{1-y}-\frac{x^{z+1}}{1-x}\bigg)\\ \end{align}$$