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.

$$ P(X \geq kY) = \sum_{y=1}^{\infty} P(X \geq kY \ | \ Y = y) P(Y = y) 
= \sum_{y=1}^{\infty} P(X \geq ky) 2^{-y} $$ 
Now 
$$ P(X \geq ky) = 1 - P(X < ky) = 1 - \sum_{r=1}^{ky - 1} 2^{-r} = 1 - 1 + (1/2)^{ky-1} = (1/2)^{ky-1} $$ 
Therefore 
$$ P(X \geq kY) = \sum_{y=1}^{\infty} 2^{-ky+1} 2^{-y} = 2\sum_{y=1}^{\infty} 2^{-y(k+1)} = 2\frac{2^{-(k+1)}}{1 - 2^{-(k+1)}} = \frac{2}{2^{k+1} - 1} $$ 
which is the answer your solution gave. 
