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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?

Edit

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.

Alex M.
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Marijn
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  • Let be $S_n$ a serie that converges to $1$. Then $kS_n$ is a serie that converges to $k$, whatever $k$ is... – Martigan Sep 15 '15 at 13:02
  • @Martigan Yes, but I want every variable to be a non-zero integer. – Marijn Sep 15 '15 at 13:06
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    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... – Martigan Sep 15 '15 at 13:10
  • @Martigan That will give me only the rational numbers, my question is if there are convergent series for every real number – Marijn Sep 15 '15 at 13:21
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    Every real number has a decimal expansion. A decimal expansion is a series built out of integers. What more could you want? – Gerry Myerson Sep 15 '15 at 13:23
  • I am afraid that the line ${ x \in \Bbb Z \mid x>1 \ \text{and} \ x<-1 }$ doesn't make much sense. Do you mean ${ x \in \Bbb Z \mid x>1 } \cup {x \in \Bbb Z \mid x<-1 }$, i.e $x \in \Bbb Z \setminus {-1, 0, 1}$? – Alex M. Sep 15 '15 at 13:29
  • @GerryMyerson Yes, you can make a series like $\frac{3}{10^0}+ \frac{1}{10^1}+\frac{4}{10^2}+\frac{1}{10^3}+\dots$ to get $\pi$, but you can't write it down in finite terms. Maybe I should have asked the question more clearly :) – Marijn Sep 15 '15 at 13:45
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    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. – pshmath0 Sep 15 '15 at 13:53
  • I think you should read this MO post on definable numbers. Your last comment sounds like you're trying to come up with something like "a number definable by a series", but there are serious foundational problems with the question "what's a definable number". – Najib Idrissi Sep 15 '15 at 13:53
  • @AlexM. Thanks, I edited it. – Marijn Sep 15 '15 at 13:55
  • @Martigan You're version to get the rationals is indeed better than what I came up with. Thanks. – Marijn Sep 15 '15 at 14:02
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    You should use digits for a natural summation that converges to a real. – Alec Teal Sep 15 '15 at 14:21
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    @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. – epimorphic Sep 15 '15 at 16:20
  • Note that $$\sum_1^{\infty}{4(-1)^n\over2n+1}$$ is usually not considered to be an expression in finite terms. I think what you want is something like $$\sum_1^{\infty}f(a_1,\dots,a_r,n)$$ where the $a_i$ are integers, and $f$ can be expressed in finite terms. But then you run striaght into the countable versus uncountable problem. – Gerry Myerson Sep 15 '15 at 23:44

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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)

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Eric Wofsey
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  • Thanks. For future reference, what would be an unambiguous way to say what I mean by "expressible in finite terms"? – Marijn Sep 16 '15 at 07:40
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    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). – Eric Wofsey Sep 16 '15 at 07:56