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A few days back a question came to my mind

What is the value of $2!!!!!!!!!!!!!!!!....$ (up to infinity)?

I feel it is 2, but one of my friends said that we can't say that for infinity.

I know it comes out to be 2 for any finite value.

But what about infinity?

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    $\begingroup$ Probably you are getting a warning because the system does not want excessive use of exclamation marks; though in this case it is acceptable. $\endgroup$
    – Bart Michels
    Sep 10, 2015 at 20:16
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    $\begingroup$ what now i get is that you can say it 2 for ---->infinity but not exactly for infinity $\endgroup$
    – Display name
    Sep 10, 2015 at 20:22
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    $\begingroup$ Notation is just notation. You can't write down all those exclamation marks, just like you can't write out in full what $\ldots$ means in $1 + 1/2! + 1/3! + 1/4! + \ldots$, but it is a convenient and well-established tradition to use shorthands of various kinds to avoid giving the details of a recursive definition. So providing you know how to give the detailed formal definition if asked and providing what you write is clear to your readers, use the notation that works for you. $\endgroup$
    – Rob Arthan
    Sep 10, 2015 at 20:39
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    $\begingroup$ I don't know a well-established meaning or definition for exclamation points extending indefinitely to the right "up to infinity". I have therefore voted to close as "unclear what you are asking". Note that double factorial $k!!$ has a commonly used meaning/definition, which is not the same as $(k!)!$. $\endgroup$
    – hardmath
    Sep 10, 2015 at 21:50
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    $\begingroup$ @hardmath Did you really vote to close this question on those grounds? I'm sad... $\endgroup$
    – Zubin Mukerjee
    Sep 10, 2015 at 22:14

7 Answers 7

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To be formal, you are perfectly entitled to define a sequence of numbers $s_n$ such that: $$ \begin{array}{rcl} s_0 &=& 2 \\ s_{n+1} &=& s_n! \end{array} $$ so that $s_n = 2! \ldots !$ with $n$ exclamation marks. But then, because $2! = 2$, you can prove by induction that $s_n = 2$ for all $n$ and this means $s_n$ tends to the limit $2$ as $n$ tends to infinity. I don't think it is harmful to think of this limit informally as $2!!!\ldots$ with a countable infinity of exclamation marks.

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    $\begingroup$ I disagree. It is particularly harmful to think about this as countably infinite number of factorials, as it lends itself to an extension that these things are "usually working out". It's easier to teach people to work correctly from the get go, and not weep for their mistakes later on. $\endgroup$
    – Asaf Karagila ♦
    Sep 10, 2015 at 22:14
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    $\begingroup$ Good answer, just wondering why you specify a "countable" infinity, what problems are there with applying induction with uncountable infinity? $\endgroup$
    – Siwel
    Sep 10, 2015 at 22:16
  • $\begingroup$ @Siwel: I wrote "countable" to keep things simple for the OP. $\endgroup$
    – Rob Arthan
    Sep 10, 2015 at 22:19
  • $\begingroup$ @AsafKaragila: I am not a professional teacher, but I did take care to give the formal explanation first. I don't see this as a question of "working correctly" but as one of thinking of useful (or at least tolerably acceptable) analogies between the finite and the infinite (no worse than the usual way we talk about sums and products of infinite sequences). $\endgroup$
    – Rob Arthan
    Sep 10, 2015 at 22:32
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    $\begingroup$ Does induction work for when treating with limits? I thought that was false. $\endgroup$
    – YoTengoUnLCD
    Sep 10, 2015 at 22:52
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You have to formally state what an infinite number of factorials is meant to be. You could define a sequence by: $$\begin{align*}a_0 &= 2 \\ a_n &= a_{n - 1}! \qquad \text{ for } n > 1\end{align*}$$

This means for example: $$a_3 = a_2! = a_1 !! = a_0 !!! \\ a_4 = a_3! = a_2!! = a_1 !!! = a_0 !!!! \\ \ldots$$

The limit of $a_n$ is what we might understand as the value of "$a_0!!!\ldots$".

It is easy to see that $a_n = 2$ for all $n$, so the limit of $a_n$ is $2$. Therefore the term "$2!!!!\ldots$" can be interpreted as $2$.

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$2!$ is just $2$. So doing $2!...!$ finitely many times produces $2$ as well.To say $2!...!$ up to infinity doesn't really make sense. What you can ask is: what is the limit of $b_n$ as $n\to \infty$, where $b_1 = 2$ and $b_n = b_{n-1}!$ for each $n$? The answer is $2$, since $(b_n)$ is a constant sequence.

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It is not as simple as some people are arguing! Consider a sequence $s_n=(2+1/n)!![n-{\rm times}]!!$ one can also say the limit of this sequence is $2!!![{\rm infinitely\; many}]$, however the result, I am sure, is divergent and is not $2$. Basically one can get any result by changing the definition.

Please give an argument before downvoting. (in case you don't know factorial is perfectly defined for non-integer values by $x!=\Gamma(1+x)$

Let me specify the most general definition: let $t_n$ is a convergent to $2$ sequence, then $$2!!!\dots\equiv \lim_{n\to\infty} t_n ![n-{\rm times}]$$ Conjecture (may be wrong) would be then that $2!!!\dots \ge 2$

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  • $\begingroup$ I'm also curious as to why this is being downvoted. It seems like a perfectly valid point. $\endgroup$
    – hasnohat
    Sep 10, 2015 at 21:52
  • $\begingroup$ Perhaps it is being downvoted because instead of providing an Answer, the point is being made that the Question is not well posed. One can agree with the latter point without agreeing with this use of the Answer box. (I am not a downvoter here.) $\endgroup$
    – hardmath
    Sep 10, 2015 at 22:01
  • $\begingroup$ @hardmath That seems like the only correct way to respond to this question, though. It's just like when people ask about $0^0$, or $0.000...01$; the thing to do is explain how to correctly use mathematical notation. Pretending that there is an answer (as some have done) is simply wrong. $\endgroup$
    – hasnohat
    Sep 10, 2015 at 22:47
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    $\begingroup$ I think the problem is that OP's expression is clearly about lots of factorials and about the number $2$. Introducing a sequence that converges to $2$ is not suggested by OPs expression. That's a complication you are introducing. I mean, no one thinks of $0+0+0+\cdots$ as $\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}$ just because $\frac{1}{n}$ converges to $0$. $\endgroup$
    – 2'5 9'2
    Sep 10, 2015 at 23:51
  • $\begingroup$ Your example is misleading, I disagree. The correct analog would be rather $\lim_{n\to \infty} \frac{1}{n} = 0$ than what you said. Or if you want $\lim_{n\to\infty}(1/n)/(2/n)$ as a resolution of $0/0$ ambiguity. $\endgroup$
    – Nikolay Gromov
    Sep 11, 2015 at 0:53
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Hmm, I guess it will just be 2.

We can continue to add faculty to the number - but the value of the faculty will remain 1; so we'll keep on adding ()*1.

(((((((2*1)*1)*1)*1)*1)*1)*1)*1...*1

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  • $\begingroup$ can we say that for infinity? $\endgroup$
    – Display name
    Sep 10, 2015 at 20:10
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    $\begingroup$ @YashMalik How would you define $2\underbrace{!!\cdots}_\infty$ if not as $\lim_{n\to\infty}2\underbrace{!!\cdots !}_n$? $\endgroup$
    – Hagen von Eitzen
    Sep 10, 2015 at 20:13
  • $\begingroup$ "faculty"???? Is this meant to be "factorial"? $\endgroup$
    – Alex M.
    Sep 10, 2015 at 20:27
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    $\begingroup$ @AlexM. Probably a German speaker, "factorial = Fakultät". $\endgroup$
    – Daniel Fischer
    Sep 10, 2015 at 21:28
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    $\begingroup$ @DanielFischer or a Dutch speaker, "factorial = faculteit" ;-) $\endgroup$
    – Bart Michels
    Sep 10, 2015 at 21:29
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Let us define $S=2!!!!!!!...$ Now, $$ S! = (2!!!!\dots)! = S\\ S=S!\\ S\in {1,2} $$

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  • $\begingroup$ Your answer gives the impression that $1$ is a plausible value. Just because $x = 2$ and $x = -2$ are solutions to $x^2 - 4 = 0$, should I worry about whether $2 = -2$? $\endgroup$
    – pjs36
    Sep 10, 2015 at 21:35
  • $\begingroup$ I love these kinds of arguments so much that I feel like there must be some way to eliminate one of the values. $\endgroup$
    – Will R
    Sep 10, 2015 at 21:55
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Just write $a_n:=2!\cdots!$ with $n$ exclamation marks. You want to compute $\lim_{n\to+\infty}a_n$. But you can easily show by induction that $a_n=2\;\forall n\in\Bbb N$, hence you're searching the limit of a constant sequence.

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