I have the following problem:
Let $X$ and $Y$ be independent random variables taking values in the positive integers and having the same mass function $P(X = k) = P(Y = k) = 2^{−k}$ for $k = 1, 2, \dots$
Find $P (X > kY)$, for a given positive integer $k$.
My solution is as follows:
$$\begin{align} P(X > ky) &= \sum_{i = 1}^\infty P(X > kY, Y = i) \ \ \text{(Joint PMF of $X$ and $Y$.)} \\ &= \sum_{i = 1}^\infty P(X > kY | Y = i) P(Y = i) \ \ \text{(By the law of total probability.)} \\ &= \sum_{i = 1}^\infty P(X > ki) P(Y = i) \ \ \text{(Since $X$ and $Y$ are independent.)} \\ &= \sum_{i = 1}^\infty \sum_{j = ki + 1}^\infty 2^{-i} 2^{-j} \ \ \text{($j = ki + 1$ since we have $X > ki$.)} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-i} 2^{-(j' + ki)} \ \ \text{(Set $j' = j - ki = 1$ to setup as geometric series.)} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-i} 2^{-j' - ki} \\ &= \sum_{i = 1}^\infty \sum_{j' = 1}^\infty 2^{-j'} 2^{-i(k + 1)} \\ &= \sum_{i = 1}^\infty 2^{-i(k + 1)} \ \ \left(\text{Since the geometric series becomes $1 \dfrac{2^{^{-1}}}{1 - 2^{-1}} = 1$}\right) \end{align}$$
The solution proceeds from where I ended as follows:
$$\begin{align} \sum_{i = 1}^\infty 2^{-i(k + 1)} &= 2^{-(k + 1)} \sum_{i = 1}^\infty 2^{-(i - 1)(k + 1)} \\ &= 2^{-(k + 1)} \dfrac{1}{1 - 2^{-(k + 1)}} \\ &= \dfrac{1}{2^{k + 1} - 1} \end{align}$$
It is not clear to me how the author made this leap in their reasoning. I would greatly appreciate it if people could please take the time to clarify this.
\begin{align} \sum_{i = 1}^\infty 2^{-i(k + 1)} &= \sum_{i = 1}^\infty 2^{-(i-1+1)(k + 1)} \\&= 2^{-(k + 1)} \sum_{i = 1}^\infty 2^{-(i - 1)(k + 1)} \\ &= 2^{-(k + 1)} \dfrac{1}{1 - 2^{-(k + 1)}} \\ &= \dfrac{1}{2^{k + 1} - 1} \end{align}
The line where they get rid of the summation sign is by identitying that it is a geometric sum of common ration $2^{-(k+1)}$ with the first term being $1$.
$$\sum_{i=0}^\infty r^i=\frac1{1-r}$$
if $|r|<1$.