# Math Help - Forumula for the product of the first 'n' numbers?

1. ## Forumula for the product of the first 'n' numbers?

I know that we are able to derive a formula for the sum of the first 'n' positive natural numbers, that sum being:

$\sum_{i=1}^n i=\frac{(n)(n+1)}{2}$

I know that we can derive this formula by writing out the sum as

$S =1+2+ \cdot \cdot \cdot \cdot + (n-1) + (n)$

and doing some algebraic manipulation. But, what if we have the following:

$\prod_{i=1}^n i = (1)(2)(3) \cdot \cdot \cdot \cdot (n-2)(n-1)(n)$

Is there a way in which we can manipulate the product expansion (simmilar to the way in which we can manipulate the summation expansion) which will cause us to arrive at a direct formula for the product of the first 'n' positive integers simmilar to

$\sum_{i=1}^n i=\frac{(n)(n+1)}{2}$

such that we'd have something like the following

$\prod_{i=1}^n = f(n)$

[p.s. also, I realize that $n!$ would denote the function I am refering to, but that sort if "notation" is not what I am refering to here. I'm asking weather there exists an explicit forumula for the product which does not have to be written in the form $(n)(n-1)(n-2)......(3)(2)(1)$ or any form that includes the ".....", but rather a precise formula simmilar to the one used for the sum of the first "n" positive integers.]

2. Why do you want to find something more complicated ? You can view it like n*(n-1)!

3. Originally Posted by mfetch22
[p.s. also, I realize that $n!$ would denote the function I am refering to, but that sort if "notation" is not what I am refering to here. I'm asking weather there exists an explicit forumula for the product which does not have to be written in the form $(n)(n-1)(n-2)......(3)(2)(1)$ or any form that includes the ".....", but rather a precise formula simmilar to the one used for the sum of the first "n" positive integers.]
I'm a little bit concerned that your question is not well-posed. What do you mean by "precise formula"?

4. The reason why we define n! to be the product of the first n positive integers is precisely because there is no simple formula for that product!

I suspect that the reason there is a nice formula for the sum but not the product is that the ordering is defined by addition (1, 1+ 1, 1+ 1+ 1, ...) and multiplication just does not "play nicely" with addition.

5. It's worth mentioning Stirling's Approximation. The fact that people bother with this may help convince the OP that coming up with a "nice" formula for $n!$ is a lofty goal indeed.

6. The product of the 'first n numbers' with n>0 is the value of the 'factorial function'...

$\displaystyle z!= \int_{0}^{\infty} t^{z}\ e^{-t}\ dt$ (1)

... when z=n ...

Kind regards

$\chi$ $\sigma$