I've come up with an example that I think can represent every non-zero rational number:

$$\sum_{n=0}^\infty \frac{k}{mx^n}$$

where $k\in \mathbb{Z}_{\neq0}$, $m \in \mathbb{N}_{\neq0}$ and $x \in \Bbb Z \setminus \{ -1, 0, 1 \}$.

And I know there are series that converge to irrational numbers like $\sqrt{2}$ and transcendental numbers like $\pi$ or $e$, that use a finite number of variables that are non-zero integers. But is there such a series for every real number?


To be more clear, I am talking about infinite series that converge to a real number, that can be expressed in finite terms. For example $\pi$ can be expressed as $$\sum_{n=0}^\infty \frac{4(-1)^{n}}{(2n+1)}$$ This would fit the bill as opposed to $$\frac{3}{10^0} + \frac{1}{10^1} + \frac{4}{10^2} + \frac{1}{10^3} + \dots$$ which would require infinitely many terms to describe.

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    $\begingroup$ Well... $S_n=\sum_{i=0}^n \frac{k}{m2^i}$ works, with $\frac km=\frac q2$, with $q$ the rational you want to reach... $\endgroup$
    – Martigan
    Sep 15, 2015 at 13:10
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    $\begingroup$ Every real number has a decimal expansion. A decimal expansion is a series built out of integers. What more could you want? $\endgroup$ Sep 15, 2015 at 13:23
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    $\begingroup$ Your sum is trivial since if we let $x=2$, we get $\sum_{n=0}^\infty\frac{1}{2^n}=2$. Hence you can choose $k$ and $m$ such that $\frac{k}{m}\cdot 2=\ell$, where $\ell\in\mathbb{Q}$ is any rational you like. $\endgroup$
    – pathfinder
    Sep 15, 2015 at 13:53
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    $\begingroup$ You should use digits for a natural summation that converges to a real. $\endgroup$
    – Alec Teal
    Sep 15, 2015 at 14:21
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    $\begingroup$ @Marijn Could you example series for $\sqrt{2}$, $\pi$, and $e$? It seems like everyone, including yours truly, is confused about exactly what kinds of series you are allowing. $\endgroup$
    – epimorphic
    Sep 15, 2015 at 16:20

1 Answer 1


No, at least for most definitions of what you mean by "expressible in finite terms". There are only countably many expressions of finite length that you can write down, and there are uncountably many real numbers, so not all of them can correspond to some expression.

(Caveat: This argument only works if the kinds of "finite expressions" you allow are limited enough that you can give a precise definition of what it means for a real number to be equal to the expression. If you interpret "finite expression" broadly enough, it turns out that this is impossible because of deep logical issues related to Gödel's incompleteness theorems. See this answer on MO, for instance)

  • $\begingroup$ Thanks. For future reference, what would be an unambiguous way to say what I mean by "expressible in finite terms"? $\endgroup$
    – Marijn
    Sep 16, 2015 at 7:40
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    $\begingroup$ You could specify exactly what operations you're allowed to use when writing down a formula for the $n$th term in the series (e.g., arithmetic operations, factorials, integer constants). Writing this out as a rigorous definition would take a bit of work and would involve some kind of inductive definition (see the discussion here, for instance). $\endgroup$ Sep 16, 2015 at 7:56

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