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Is there a sequence of rational numbers $a_n$ such that $a_1 e^{-1} + a_2 e^{-2} + cdots = 1$?

Mathematics Asked by NiloS on December 22, 2020

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 $x$.

One Answer

Given an arbitrary real number $x in (0,1)$, we can construct a sequence of rational numbers $(a_n)_{n = 1}^{infty}$ such that $displaystylesum_{n = 1}^{infty}a_nx^n = 1$ as follows:

Let $a_1 = leftlfloordfrac{1}{x}rightrfloor$. For each integer $n ge 2$, let $a_n = dfrac{1}{n}leftlfloor dfrac{n}{x^n}left(1-displaystylesum_{k = 1}^{n-1}a_kx^kright) rightrfloor$.

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 $displaystylesum_{k = 1}^{n}a_nx^n$ does not exceed $1$. With this construction, $left|1-displaystylesum_{k = 1}^{n}a_kx^kright| le dfrac{1}{n}$ for all integers $n ge 1$, and thus, $displaystylesum_{k = 1}^{infty}a_kx^k = 1$.

Correct answer by JimmyK4542 on December 22, 2020

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