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Let $X$ and $Y$ be independent random variables such that $$\mathbb{P}(X=n)=\mathbb{P}(Y=n)=\frac{1}{2^n}\quad \quad (n\in\mathbb{N})\;\text{.}$$ Find the probability that $X$ divides $Y$.

Here's what I tried to do: $$\lbrace X \text{ divides } Y\rbrace=\bigcup_{n=1}^\infty(\lbrace X=n\rbrace\cap\lbrace n \text{ divides } Y\rbrace)=\bigcup_{n=1}^\infty\bigcup_{k=1}^\infty(\lbrace X=n\rbrace\cap \lbrace Y=kn\rbrace)\;\text{.}$$ I wrote the probability as a series and after some calculations I got $$\mathbb{P}(X\text{ divides } Y)=\sum_{k=2}^\infty\frac{1}{2^k-1}\;\text{.}$$ I don't know how to find the sum of this series. Is there maybe some kind of trick that I can use to avoid this series?

fred2015
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  • The sum is also $$\sum_{n=1}^\infty\frac{\nu_2(n)-1}{2^n}$$ where $\nu_2(n)$ denotes the number of divisors of $n$, but apart from that... – Did Jun 07 '17 at 08:07
  • Interesting. Hard to get intuition for the answer...numerically your sum is about $.6067$ and there is a $\frac 13$ chance that $Y>X$ so we know the answer is at most $\frac 23$. Of course $P(X=1)=\frac 12$ so we know the answer exceeds $\frac 12$ and there is a $\frac 1{12}$ probability that $X=Y\neq 1$ so the answer exceeds $\frac 7{12}$. That is very close to your answer.... – lulu Jun 07 '17 at 09:48
  • You can't really avoid this series in this case, but people have asked about this sum (or closely related ones) before: https://math.stackexchange.com/questions/1978310/find-sum-k-1-infty-frac12k1-1/1978329 , https://math.stackexchange.com/questions/203125/can-this-series-be-expressed-in-closed-form-and-if-so-what-is-it – Michael Lugo Jun 07 '17 at 12:35

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