A necessary condition for a series of rational numbers to be irrational? Suppose $(q_{n})_{n\in\mathbb{Z}_{\gt 0}}$ is a decreasing sequence of positive rational numbers such that $Q:=\displaystyle{\sum_{n>0}q_{n}}$ is finite. Let's denote by $n_{i}$ and $d_{i}$ the numerator and the denominator of $q_{i}$ such that $(n_{i},d_{i})=1$. If $i\ne j\Rightarrow (n_{i}d_{i},n_{j}d_{j})=1$, is $Q$ necessarily irrational?  

No. You can use a greedy algorithm to construct a counterexample. For example, let $n_1/d_1 = 2/3$. Given $n_1/d_1,\dots,n_k/d_k$ whose sum is $S_k$, choose a rational number $n_{k+1}/d_{k+1}$ in the interval $[\frac12(1-S_k),\frac23(1-S_k)]$ whose numerator and denominator are relatively prime to $n_1\dots n_kd_1\dots d_k$. Then $\sum_{k=1}^\infty n_k/d_k$ equals $1$ by construction.
