# 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 $$x$$.

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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