Question: Let $\{x_n\}_{n\ge 0}$ be a sequence of positive reals for which $\sum_{n\ge 0}x_n$ converges. Does there always exist a sequence $\{y_n\}_{n\ge 0}$ of rationals such that $0<y_n<x_n$ for all $n\ge 1$, and $\sum_{n\ge 0}y_n$ converges to a rational number?
If not, can you give a counterexample $\{x_n\}_{n\ge 1}$?
Motivation: Let $\{a_n\}_{n\ge 1}$ and $\{b_n\}_{n\ge 1}$ be sequences of positive integers, and suppose that $\alpha:=\sum_{n\ge1}\frac{a_n}{b_n}$ converges. We'd like to place some conditions on the sequences $\{a_n\}_{n\ge 1}$ and $\{b_n\}_{n\ge 1}$ and conclude that $\alpha$ is irrational by imitating the classical proof that $e$ is irrational.
Suppose that $\alpha = p/q$, then for all $m\ge 1$, we have that $$ 0 < q\cdot \operatorname{lcm}(b_1,\ldots,b_m)\cdot\left(\alpha-\sum_{n=1}^m\frac{a_n}{b_n}\right)=q\cdot\operatorname{lcm}\left(b_1,\ldots,b_m\right)\sum_{n=m+1}^{\infty}\frac{a_n}{b_n} $$ is a positive integer. So if $$ \lim_{m\to \infty}\operatorname{lcm}(b_1,\ldots,b_m)\sum_{n=m+1}^\infty\frac{b_n}{a_n}=0, $$ we obtain a contradiction for $n$ large enough. Hence, $\alpha$ must be irrational in this case.
What we need for this argument to work, is for $a_n/b_n$ to decrease fast enough, where 'fast enough' is determined by how quickly $\operatorname{lcm}(b_1,\ldots,b_n)$ grows as a function of $n$. I'm asking if we really need a second condition besides $a_n/b_n$ decreasing fast.
Sure and you can construct it explicitly. Let $x=\sum x_i\in \mathbb{R}$ and let $0<y<x$ be a chosen rational number. Let $$ \frac{x_0}{x}y< y_0< \min\{y,x_0\}. $$ Now 'renormalize' if you will by replacing $x$ with $x-x_0$ and $y$ with $y-x_0$ and define $y_1$ just as we defined $y_0$ above. Repeat.
The result is a sequence $y_i$ of rational numbers so that $$ \frac{y}{x}\sum_{i=0}^n x_i<\sum_{i=0}^n y_i <y $$ The squeeze theorem proves convergence of $S_n:=\sum_{i=0}^n y_i \to y$