What is $2!!!!!!!!!!!!!!!!!!!!$................. (up to?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

$2!$ 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.

Answer 4

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 $2$.

Answer 5

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 $2!!!\dots \ge 2$

Answer 6

Let us define $S=2!!!!!!!...$ Now, $$ S! = (2!!!!\dots)! = S\\ S=S!\\ S\in {1,2} $$

Answer 7

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.