How do I figure this one out? $p\in (0,1) \sum_{k\in Uneven}(1-p)^{k-1}p$ How do I figure this one out? $$p\in (0,1) \sum_{k\in Uneven}(1-p)^{k-1}p; Uneven=\\{1,3,5,7...\\}$$ I know that if $k\in \mathbb N$ then this sum would be equal to one, because this is like the geometric distribution. But what about only over uneven numbers? Just $\frac{1}{2}?$

Think about why it is equal to 1 for $k \in \mathbb N$ and extend it. Let $p \in (0,1)$. You want to find

$$\sum_{k\in \\{1,3,5,...\\}}^\infty(1-p)^{k-1}p$$ which is equivalent to just summing $p(1-p)^k$ for $k \in \\{0,2,4,6,...\\}$ which is equivalent to:

$$p\sum_{k=0}^\infty(1-p)^{2k}$$

This sum is equal to the infinite sum of a geometric series with $r = (1-p)^2$ and so you have

$$ \begin{split} \text{Sum} &= p\sum_{k=0}^\infty(1-p)^{2k} \\\ & = p\frac{1}{1-(1-p)^2} \\\ &= \frac{p}{2p-p^2} \\\ &= \frac{1}{2-p} \end{split} $$