Calculating a sum sigma, with minimum $\sum_{i=1}^{\min(n,k-1)}\frac{1}{(1-p)^i}$

Question

How do I calculate this sum:

$$\sum_{i=1}^{\min(n,k-1)}\frac{1}{(1-p)^i}$$

It is like a geometric sum, but it has the minimum which I do not know how to deal with.

I got this sum while calculating two random variables $X+Y$, such that $X$~$U(1,n)$ , $Y$~$G(p)$, $n \in \mathbb{N}, 0<p<1$.

$$\begin{align}P(X+Y=k) &= \sum_{i=1}^{\min(n,k-1)} P(X=i,Y=k-i) \\[1ex] &=\sum_{i=1}^{\min(n,k-1)} P(X=i) \cdot P(Y=k-i) \\[1ex] &=\sum_{i=1}^{\min(n,k-1)} \frac{1}{n} \cdot (1-p)^{k-i-1}p\\[1ex]&=\frac{(1-p)^{k-1}p}{n} \cdot \sum_{i=1}^{\min(n,k-1)} \frac{1}{(1-p)^i} \\[1ex] &\end{align}$$

Thanks a lot!

Answer 1

If you dont want to include the $\min(a,b)$ function then you could rewrite the sum as follows: $$\begin{align*}\sum_{i=0}^{\min(n,k-1)}\frac{1}{(1-p)^i}&=\min\bigg(\sum_{i=0}^{n}\frac{1}{(1-p)^i},\sum_{i=0}^{k}\frac{1}{(1-p)^i}\bigg)\\&=\frac{\sum_{i=0}^{n}\frac{1}{(1-p)^i}+\sum_{i=0}^{k}\frac{1}{(1-p)^i}-\bigg|\sum_{i=0}^{n}\frac{1}{(1-p)^i}-\sum_{i=0}^{k}\frac{1}{(1-p)^i}\bigg|}2\end{align*}$$

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Note: $\sum_{i=0}^{\min(n,k-1)}\frac{1}{(1-p)^i}=\min\bigg(\sum_{i=0}^{n}\frac{1}{(1-p)^i},\sum_{i=0}^{k}\frac{1}{(1-p)^i}\bigg)$ is true because $p-1>0$.