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Is it possible to express all transcendental numbers (and more generally all real numbers $\in \mathbb R$) as the sum of an infnite series of the form

$$\sum_{n=0}^{∞} \frac{p(n)}{q(n)}$$ where $p(n)$ and $q(n)$ are polynomials with rational coefficients (we assume that the degree of $p$ and $q$ are integers and that they differ by at least $2$, which is required for convergence ).

My approach to this problem would be to show that the set of algebraic numbers forms a countable set, and that the set of rational functions $\frac{p(n)}{q(n)}$ would also be countable, so the transcendental numbers would form a countable set --- but they are known to form an uncountable set, a contradiction. This would indicate that not all transcendental can be expressed by the following series.

Any ideas for a formal proof? Was my approach correct? Thanks for help.

J. Linne
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    correct, since you are using rational coefficients, there will only be countably many rational functions $\frac{p(n)}{q(n)}$. If you fix the degree of $p$ say $m$, and also fix the degree of $q$, say $k$ then as a first step show that there are only be countably many rational functions $\frac{p(n)}{q(n)}$ (for these fixed degrees). As a next step use that there are only countably many choices for $m$ and $k$, so you will be using that the union of countably many countable sets is countable. – Mirko Sep 11 '21 at 05:47
  • What is an explicit example of a transcendental number that does not have such a representation? – Sidharth Ghoshal Nov 29 '23 at 03:17

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Yes, your suggestion works: more generally, given any countable set $\mathcal{F}$ of functions, only countably real numbers can have the form $\sum_{n\in\mathbb{N}}f(n)$ for some $f\in\mathcal{F}$ so "most" real numbers won't be expressible in such a way (here $\mathcal{F}$ is the set of rational functions with rational coefficients).

Noah Schweber
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