Every positive real number $x_1$ is, in one and only one way, expressible in the form: $$x_1 = a_1 +{a_2\over2!}+{a_3\over3!}+\cdots$$ where $a_n$ is a non-negative integer with $a_n \le n-1$ for $n > 1$ subject to the condition of not being $=n-1$ for every $n$ after a definite $n_0$. If $x_1$ is rational, and only then, the series terminates. For rational $x_1$ there's already an answer here Every positive rational number $x$ can be expressed in the form $\sum_{k=1}^n \frac{a_k}{k!}$ with $ a_k≤ k − 1$ for $k ≥ 2$ . but I don't understand the uniqueness part. How to generalize this to real numbers?
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1Can you elaborate on what you do not understand about the uniqueness part? E.g. Which sentence did you find confusing? Which sentence do you not agree with? – Calvin Lin Mar 14 '20 at 01:05
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@Calvin Lin I don't see how the hint given by Brian Scott can be useful – Peanut Mar 14 '20 at 10:23
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@CalvinLin also can you elaborate your answer about uniqueness? – Peanut Mar 14 '20 at 10:55
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1For $a_1 = 0$, what is the largest value you can get with a sum up to the $1/n!$ term? That is, what is $$\sum_{k=2}^n \frac{k-1}{k!}\text{?}$$ So if $x = a_1 + a_2/2!+\ldots$ and $x = b_1 + b_2/2!+\ldots$, is it possible for $b_1 > a_1$? – Paul Sinclair Mar 14 '20 at 14:46
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if the sum is finite no, for $a_1$ is the integer part of $x$ and similarly $b_1$. In fact one can find a formula for $a_k$. The problem is when the sum is infinite. – Peanut Mar 14 '20 at 15:03