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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 $0$.

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  • $\begingroup$ Did you mean $y^z$ or $y^i$? $\endgroup$ – Frpzzd Nov 11 '18 at 0:21
  • $\begingroup$ Thanks, corrected $\endgroup$ – Y.L Nov 11 '18 at 0:25
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Yes, just use the formula $$\sum_{n=0}^\infty \alpha^n=\frac{1}{1-\alpha}$$ for $|\alpha|<1$. We have that $$\begin{align} \sum_{i=z}^\infty y^i \frac{1-(x/y)^{i+1}}{1-x/y} &=\frac{1}{1-x/y}\sum_{i=z}^\infty y^i-\frac{x}{y-x}\sum_{i=z}^\infty x^i\\ &=\frac{1}{1-x/y}\frac{y^z}{1-y}-\frac{x}{y-x}\frac{x^z}{1-x}\\ &=\frac{y^{z+1}}{(1-y)(y-x)}-\frac{x^{z+1}}{(1-x)(y-x)}\\ &=\frac{1}{y-x}\bigg(\frac{y^{z+1}}{1-y}-\frac{x^{z+1}}{1-x}\bigg)\\ \end{align}$$

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  • $\begingroup$ Thank you so much! $\endgroup$ – Y.L Nov 11 '18 at 0:37

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