Let $e$ be the base of the natural logarithm. Is there a sequence of rational numbers $a_n$ such that
$$ \frac{a_1}{e} + \frac{a_2}{e^2} + \frac{a_3}{e^3} + \cdots = 1 $$
More generally, under what conditions on a given a irrational number $0 < x < 1$, can we find a sequence of rational number $a_n$ such that $\sum_{n=1}^{\infty} a_n x^n = 1$. I have been able to construct algebraic irrational number for which I can find a suitable $a_n$ but in general I am not sure if this is possible for an arbitrary irrational $x$.
Given an arbitrary real number $x \in (0,1)$, we can construct a sequence of rational numbers $(a_n)_{n = 1}^{\infty}$ such that $\displaystyle\sum_{n = 1}^{\infty}a_nx^n = 1$ as follows:
Let $a_1 = \left\lfloor\dfrac{1}{x}\right\rfloor$. For each integer $n \ge 2$, let $a_n = \dfrac{1}{n}\left\lfloor \dfrac{n}{x^n}\left(1-\displaystyle\sum_{k = 1}^{n-1}a_kx^k\right) \right\rfloor$.
Note that once $a_1,\ldots,a_{n-1}$ are chosen, $a_n$ is the largest rational number of the form $\dfrac{m}{n}$ such that $\displaystyle\sum_{k = 1}^{n}a_nx^n$ does not exceed $1$. With this construction, $\left|1-\displaystyle\sum_{k = 1}^{n}a_kx^k\right| \le \dfrac{1}{n}$ for all integers $n \ge 1$, and thus, $\displaystyle\sum_{k = 1}^{\infty}a_kx^k = 1$.