TITLE: Proof that the following sum is bounded above.
QUESTION [6 upvotes]: Let  $i_1,  i_2 , ... , i_n, ...$ be a sequence of positive integers such that,
a) No $i_n$ is a prime.  
b) For all pairs of distinct positive integers, $m$ and $n$, the pair of integers $i_m$ and $i_n$ are relatively prime.  

Show that $\frac{1}{i_1}  + \frac{1}{i_2}+ ... + \frac{1}{i_n} + ...$ is bounded above by some
  finite real number.


I know that all $i_k$ are $> 1$ for every $k$, and $q_k$ the smallest prime that divides $i_k$(Thanks Andre for this hint).  
I try several ways (more calculus related, infinite series) but all my attempts end in a failure. Therefore I can't find such a finite real number that its bounded the sum of my problem.
For me is a very hard problem, and I would really appreciate if I receive help solving this exercise, and then been able to understand the path of thinking of how to solve such a kind of problem. 
Thanks again community.

REPLY [3 votes]: Consider the least prime $p$ dividing $i_k$. Then since $i_k$ is not prime we have that $i_k\geq p^2 $. Now, using the fact that the $i_k$ are relatively prime we see that a prime $p$ can divide at most one $i_k$, and so $$\sum_{i=1}^\infty \frac{1}{i_k}\leq \sum_{p}\frac{1}{p^2}\leq \sum_{n=1}^\infty \frac{1}{n^2}=\frac{6}{\pi^2}$$