Question

I just thought of some problem:

Find out the value of:

\(1!*2!*3!*...*n!\)

Answer 1

I believe this is called the "superfactorial"

Answer 2

I tried simplifying it:

\(1!*2!*3!*...*n!\)
\[\prod_{i = 1}^{n} (n+1-i)!\]
\[\prod_{i=1}^n (i)^{n+1-i}\]

But how to solve it after this ?

Any ideas ?

Answer 3

sorry no idea about that @kirbykirby , it was just some random question that came in my mind.

Answer 4

I don't think it simplifies any more

Answer 5

Well I think that's the end of another idea :(

Answer 6

Thanks @kirbykirby

Answer 7

How about this one:

\[\large{\prod_{k=1}^{n}(k!)^{\sum_{i=1}^{k}(i!)}}\]

Answer 8

There may be some weird way of writing it in another way though But I'm not aware of it. I only came across the superfactorial once so I don't know any nice properties for it

Answer 9

Any idea ?

Answer 10

would it not simply be \(\prod_{k=1}^n k!\) ?

Answer 11

No I don't think so.

Answer 12

Lets change it into this:

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
How about this one:

\[\large{\prod_{k=1}^{n}(k!)^{\large{\sum_{i=1}^{k}(i*i!)}}}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 13

This one is much easier I think:

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
Lets change it into this:

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
How about this one:

\[\large\color{red}{\prod_{k=1}^{n}(k!)^\color{blue}{\large{\sum_{i=1}^{k}(i*i!)}}}\]

\[= \large\color{red}{\prod_{k=1}^{n}(k!)^\color{blue}{\large{(k+1)!-1}}}\]

Answer 14

Now any idea how to continue this ?

Answer 15

What do you think @kirbykirby

Answer 16

I used this property:

\[\large{\sum_{i=1}^{k} (i!)} = (k+1)! - 1 \]

Answer 17

@mathmale

Answer 18

I don't see why \[\prod_{k=1}^n k!\] is not appropriate. Maybe k! should be in parentheses to be clear. \[ \prod_{k=1}^n (k!)\] .

Answer 19

There was no problem in that. I was thinking that you were telling me the net solution for the problem.

Sorry about that. :)

Answer 20

Oh you were giving different problem..

Answer 21

sorry I didn't realize

Answer 22

I am getting this on further simplification:

\[\large{\prod_{k=1}^{n} (k)^{\sum_{i=1}^{n}[(i+1)!-1]}}\]

Answer 23

Lets solve the power first. That will make the problem less messy.

Answer 24

Okay, I have,

\[\large{\sum_{i=1}^{n}[(i+1)!-1]}\]

Answer 25

I believe I cracked it

Answer 26

I'm not aware of that identity, and it doesn't seem to work
\[\large{\sum_{i=1}^{k} (i!)} = (k+1)! - 1 \]
say k = 4:
LHS
1! + 2! + 3! + 4! = 33

RHS:
5! - 1 = 119

Answer 27

Oh sorry I meant this:

\[\large{\sum_{i=1}^{k} (i*i!)} = (k+1)! - 1 \]

Answer 28

oh ok that makes sense

Answer 29

Sorry about that

Answer 30

Are you looking for a closed-form solution

Answer 31

Well I am frankly looking for the most simplified form

Answer 32

Solving the power:


Okay, I have,

\[\large{\sum_{i=1}^{n}[(i+1)!-1]}\]

\[=\large{\sum_{i=1}^{n}[(i+1)!]-n}\]


Again stuck here.

How to find the sum of factorials?

Answer 33

I don't think this is a simple solution: http://mathworld.wolfram.com/FactorialSums.html

Answer 34

Yeah but that was if n was not a natural number

Answer 35

That's why gamma function was used.

Answer 36

http://math.stackexchange.com/questions/451604/can-the-factorial-function-be-written-as-a-sum

This one is much better

Answer 37

But I don't think its going to raise our spirits even a small bit.

Answer 38

I think the best thing we can say is just leave it.

Its really frustrating.

Answer 39

Another one of my idea into the bin. :D

Answer 40

but if you are doing a summation from 1 to n in sigma notation, then it should be natural numbers. Otherwise you would be doing some integral if you could allow any number over an interval

Answer 41

but I agree dealing with factorials is not obvious

Answer 42

Yeah...

I think I am going to change my ideas to calculus. At least, it doesn't have such things.

On second thought, I am not so sure about it too. But, I will surely give it a try.

I remember one thought:

Always try before giving up. So that you can say that at least I got defeated after trying.

Answer 43

Well thanks for sticking with me @kirbykirby . I really appreciate your help.

Good day.

Answer 44

well good effort on your part, maybe you will discover a nice formula sometime :)

Answer 45

May be...