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Does every convergent infinite series have a closed-form value?

I apologize if this question seems totally crazy to some of you. There's a ton of series that converge, but only a fraction of them have a closed-form value, so how can I even ask such a question? Isn't it obvious that the answer is no?

Well, not so fast. My intention with this question is more like "Does the possibility exist that every convergent infinite series have a closed-form value?" or have we proven that it's impossible? A "simple" way to prove this would be to find an infinite convergent series and then show that it would be impossible for it to have a closed-form value. But how would you do that?

Sure, there is an enormous amount of series that converge, for which there is not a known closed-form value, e.g. one of the series I asked about earlier, and one may use this as an argument for the answer no, but that's not enough for me.

I hope this provokes some discussion and that you don't think this questions is completely dumb.

Casimir Rönnlöf
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    How are you defining a "closed form value"? There are "uncomputable numbers", say, just by counting arguments. If you pick one between $0$ and $1$, write it as $\alpha= .a_1a_2a_3\cdots$ then $\alpha = \sum_{i=1}^{\infty} a_i10^{-i}$. What sort of closed form would you want to get for that? – lulu May 04 '20 at 17:39
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    Let $x$ be any non-closed form real number between $0$ and $1$ (define "non-closed form" however you wish, as long as such a number actually exists). Let $0.x_1x_2x_3\ldots$ be the decimal expansion of $x.$ Then $x$ is the sum of the following infinite series: $\frac{x_1}{10} + \frac{x_2}{100} + \frac{x_3}{1000} + \cdots$ – Dave L. Renfro May 04 '20 at 17:39
  • @lulu I think the Wikipedia article, pretty much sums up my definition of closed form: "In mathematics, a closed-form expression is a mathematical expression expressed using a finite number of standard operations. It may contain constants, variables, certain "well-known" operations (e.g., + − × ÷), and functions (e.g., nth root, exponent, logarithm, trigonometric functions, and inverse hyperbolic functions), but usually no limit, differentiation or integration". I don't know how I didn't think of your and Dave's example, I'm sorry for that. Cont. in next comment. – Casimir Rönnlöf May 04 '20 at 17:55
  • @lulu But still, how can we ever be sure, e.g. in your example that there isn't a closed form for it? (some combination of pi's whatever idk). – Casimir Rönnlöf May 04 '20 at 17:56
  • Such an expression would make $\alpha$ computable. – lulu May 04 '20 at 17:57
  • With your definition, there are only countably many closed form expressions but there are obviously many more series than that. – lulu May 04 '20 at 17:59
  • @lulu Pardon me, uncountable numbers are new to me, but do we know there exists uncountable numbers? – Casimir Rönnlöf May 04 '20 at 18:00
  • By counting. All the computations you can think of can be specified by countably many parameters (there are only countably many programs that can be written in a given Universal Turing Machine, say). But there are uncountably many real numbers, so there must be real numbers which can not be the output of any such computation. – lulu May 04 '20 at 18:02
  • You mean "uncomputable", I assume. There are uncountably many reals by the standard diagonalization argument. – lulu May 04 '20 at 18:04
  • @lulu Yeah, sorry, I wrote wrong. Thank you for your helpful comments. I think Matt Samuel's answer sums up this discussion quite well. Thank you once again. – Casimir Rönnlöf May 04 '20 at 18:05
  • @DaveL.Renfro Sorry for not answering you directly, I think Lulu and you had pretty similar comments. – Casimir Rönnlöf May 04 '20 at 18:08
  • Should say: there is a question you could ask along these lines. Namely: "suppose we have a program which generates a sequence $a_i$ and that $A=\sum a_i$ converges. Must there be a closed formula for $A$? " That one shuts down the "uncomputable $\alpha$" example since in that case there is no program that generates the $a_i$. – lulu May 04 '20 at 18:08
  • Note: The answer to the variant I proposed is certainly No (I presume), but in this case I don't immediately see how to prove it, – lulu May 04 '20 at 18:09
  • @lulu Yeah I think this is more of what I want answered, is this unsolved or? – Casimir Rönnlöf May 04 '20 at 18:10
  • @lulu Ok, I'll see if I ask it again tomorrow or something I'm too tired for it now XD. Thank you. – Casimir Rönnlöf May 04 '20 at 18:11
  • Oh, I think it should be solvable, as soon as you specify precisely what you mean by closed value. I expect that one could construct a (computable) number algebraically independent of all the numbers reachable by closed form. Something like that. – lulu May 04 '20 at 18:12
  • @lulu Yeah... I have to dive a bit more into the topic (might adk the question later). If I ask it, is the Wikipedia enough to specify what I mean by closed form? Or should I use something else? – Casimir Rönnlöf May 04 '20 at 18:19
  • I'd say the definition was too vague to be of much use (but I didn't think hard enough about it so I might be mistaken). Later in that article they discuss three subfields of the complex numbers that might be good contenders for "closed form numbers". I would study each of these. – lulu May 04 '20 at 18:20
  • @lulu Yep, thanks. – Casimir Rönnlöf May 04 '20 at 18:21
  • You might find this article interesting. – lulu May 04 '20 at 18:36
  • @lulu: Incidentally, the issue with the decimal expansion (and other) examples is that we're using "nice numbers" (e.g. rational numbers) to approach a "not nice number" by doing so in "not nice ways". Note that the distances between the nice approach numbers and the not nice number are themselves not nice numbers (can't really avoid this if "nice numbers" are to have nice arithmetic closure properties). Also, the "pattern of approach" (e.g. the successive differences between the approach numbers) is probably not nice in certain ways, or at least no effort towards this aspect is being made. – Dave L. Renfro May 05 '20 at 09:20
  • @DaveL.Renfro Well, I agree that the "manner of approach" is crucial. I assume (possibly incorrectly) that things like Liouville numbers (in the approximation sense, not in the, new to me, closed form sense) are not in the closed form fields. $\sum 10^{-n!}$ say. But I haven't looked closely at the various definitions of the closed form fields. – lulu May 05 '20 at 10:11

2 Answers2

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I think the definition of "closed-form value" should include that it is expressible as a finite combination of some known constants via some algebraic (or transcendental, but still elementary) formula. Therefore, a series that converges to a number that is not computable does not have a closed form value. Since most real numbers are not computable, most series do not converge to a closed form value.

A number that is not computable is one where there is absolutely no computer program that will output all of the digits of the number, even given infinite time. In fact, since there are only countably many computer programs, only countably many real numbers are computable, even though it is, by definition, difficult to produce even one that is not. Any number with a closed form expression would be computable, and we could compute it by applying the formula.

Matt Samuel
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    I interpret the original question as - can there be a closed form convergent series that converges to a limit that has no closed form. I.e., the question is just about the [convergent] series that we can actually write down as a sum. For this (more natural) question, this counting argument does not hold - as there are only countably many such series. – Ofek Shilon Nov 04 '21 at 09:24
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Depends on what you understand by "closed form". If you mean "expressible in a finite number of (give a list of functions and constants)", absolutely not. For a few examples, the values of Riemann's zeta function:

$\begin{align*} \zeta(s) = \sum_{n \ge 1} n^{-s} \end{align*}$

are known only for even values of $s$, their values for odd values of $s$ are a complete mystery.

Simpler, even: It is easy to see that there the set of convergent series is non-denumerable, even if you restrict yourself to rational terms; but the set of finite formulas made up of algegraic numbers, a handful of "well-known" trascendentals, and a finite set of functions is denumerable. There must be series with no "closed form".

vonbrand
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