Probability, Discrete random variables 
Let $X$ and $Y$ be independent random variables, taking values in the positive integers and having the same mass function $f(x)=2^{-x}$ for $x=1,2,...$ .Find  $P(X\geq kY)$, for a given positive integer

I did: $\displaystyle P(X\geq kY)=P(Y\leq X/k)=\sum_{r=1}^{\infty}P(Y\leq r, X=rk)=\sum_{r=1}^{\infty}P(Y\leq r)P(X=rk)$
$P(Y\leq r)=\displaystyle\sum_{u=1}^{r}\frac{1}{2^{u}}= 1-\frac{1}{2^{r}}$
$P(X\geq kY)=\displaystyle \sum_{r=1}^{\infty} (1-\frac{1}{2^{r}})\frac{1}{2^{rk}}=\frac{2^k}{(2^k-1)(2^{k+1}-1)}$ 
But the solution says $P(X\geq kY)=\displaystyle\frac{2}{2^{k+1}-1}$ and it's solved by doing
$\displaystyle P(X\geq kY)=\sum_{r=1}^{\infty}P(X\geq kr,Y=r)$ i think that this approach can't be so far from mine but results are different. Am i missing something? Thanks beforehand.

To understand your mistake, fix $k=2$. In the case you iterate over $X=2 r$, thus missing all the odd numbers. For $Y\leq X/k$, $X$ must be conditions to be $Y\leq q$ and $X=q*r+s$ where $s$ runs from $0$ to $k-1$.
