Proof that the following sum is bounded above. 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.
 A: Throughout the text, I will use tags like $\color{green}{[h]}$ to cite "references" in the same text.
First of all, I'm going to use the fact that the series $\sum1/n^2$ converges (in fact the series $\sum1/n^p$ converges for $p>1$ and diverges for $p\leqslant1$). This can be shown in many ways (see here).
Now, it is absolutely necessary that $i_k>1$ for all but finitely many $k$'s, for, otherwise the sequence $\{i_n\}$ would have an infinite number of $1$'s and so clearly the series $\sum1/i_n$ would diverge. 
Let $N$ be the largest natural number $k$ such that $i_k=1.$ Then $i_n>1$ for all $n>N.$ Thus, for each $n>N,$ $i_n$ is divisible by some prime $q.$ Now, for each $n>N,$ let $p_n$ be the least prime number dividing $i_n$ $^\color{red}{[0]}$. 
Since no $i_k$ is prime, it follows that $i_n$ is composite for each $n>N$ so that $i_n$ is the product of (at least) two primes $^\color{blue}{[1]}$.
Now let's write (for each $n>N$) $i_n=q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_m^{\alpha_m}$ (the prime factorization of $i_n$), where $m$ is at least $2$ $\color{blue}{[1]}$ and $q_1<q_2<\cdots<q_m$ (this means that $q_1$ is the least prime factor of $i_n,$ that is, $q_1=p_n$ $\color{red}{[0]}$). Then 
$$
\begin{aligned}
i_n&=q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_m^{\alpha_m}\\\\&>\underbrace{q_1^{\alpha_1}q_1^{\alpha_2}\cdots q_1^{\alpha_m}}_{\text{$m$ terms}}\\\\&\geqslant\underbrace{q_1q_1\cdots q_1}_{\text{$m$ times}}\\\\&=(q_1)^{m}\\\\&=(p_n)^m\\\\&\geqslant(p_n)^2
\end{aligned}
$$
and hence $\dfrac{1}{i_n}<\dfrac{1}{(p_n)^2}$ for each $n>N$ $^\color{green}{[2]}$.
Now, it is easy to see that for each positive integer $k$ we have $k\leqslant t_k$ where $t_k$ denotes the $k$-th prime and hence $\dfrac{1}{(t_k)^2}<\dfrac{1}{k^2}.$
Therefore, for each $n>N$ we have 
$$
\begin{aligned}
\sum_{k=1}^n\dfrac{1}{i_k}&=\sum_{k=1}^N\dfrac{1}{i_k}+\dfrac{1}{i_{N+1}}+\dfrac{1}{i_{N+2}}+\cdots+\dfrac{1}{i_{n}}\\\\&<\sum_{k=1}^N\dfrac{1}{i_k}+\dfrac{1}{(p_{N+1})^2}+\dfrac{1}{(p_{N+2})^2}+\cdots+\dfrac{1}{(p_n)^2}\\\\&\leqslant\sum_{k=1}^N\dfrac{1}{i_k}+\dfrac{1}{(t_1)^2}+\dfrac{1}{(t_2)^2}+\cdots+\dfrac{1}{(t_{n})^2}\\\\&\leqslant\sum_{k=1}^N\dfrac{1}{i_k}+\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}\\\\&<\sum_{k=1}^N\dfrac{1}{i_k}+\sum_{k=1}^\infty\dfrac{1}{n^2}
\end{aligned}
$$
and hence $\sum\limits_{k=1}^n\dfrac{1}{i_k}<\sum\limits_{k=1}^N\dfrac{1}{i_k}+\sum\limits_{k=1}^\infty\dfrac{1}{n^2}$ for all $n\geqslant1$ which implies that $$\sum\limits_{n=0}^\infty\dfrac{1}{i_n}\;\leqslant\;\sum\limits_{k=1}^N\dfrac{1}{i_k}+\sum\limits_{k=1}^\infty\dfrac{1}{n^2}$$ and since the LHS of the inequality is a finite real number, we are done.
A: 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 $$\sum_{i=1}^\infty \frac{1}{i_k}\leq \sum_{p}\frac{1}{p^2}\leq \sum_{n=1}^\infty \frac{1}{n^2}=\frac{6}{\pi^2}$$ 
