Can you write something like $3\cdot5\cdot7\cdots(2n+1)$ as $(2n+1)!$?

Question

I have been working through some infinite series in calculus I, and there have been a lot of expressions of the form $a\cdot b\cdot c\cdots(zn+x)$. It usually seems to work if I simplify them to $(zn+x)!$. In addition to the example I gave in the question, here are a couple more:

1. $2\cdot3\cdot4\cdots(n+1) = (n+1)!$
2. $1\cdot2\cdot3\cdots(2n-1) = (2n-1)!$

However, my professor said that this was incorrect and that she had an explanation for why, but for some reason, it slipped her mind. It seems like my textbook does not simplify them like I do either, but I still do not understand what is mathematically wrong with doing it like that.

Any help would be much appreciated!

-Isaac

Edit:

Thanks for your comments. I see that I am clearly overthinking this, and I'm sure that something will click into place once I read through what you all said several times more. However, I still don't really get it. Isn't

$(n+1)! = (n+1)\cdot(n)\cdot(n-1)\cdot\cdots\cdot(n-(n-1))$?

How is this not the same as

$(n+1)! = 2\cdot3\cdot4\cdot\cdots\cdot(n+1)$?

Do I have to specify that n is an integer greater than 1 for it to work, or is there something more fundamentally wrong?

I'm not trying to waste anyone's time; this is honestly something that I am confused about. I enjoy my math class, and I want to really get what I am learning.

Also, thanks for posting that MathJax tutorial. I was trying to figure it out earlier, but I think I've got it now.

One last note: my original post had a typo in example 2. It is fixed now.

Answer 1

I think you are looking for the product notation.

$$\prod_{i=1}^n (2i-1)= 1\cdot 3\cdot 5 \ldots (2n-1)=\frac{(2n-1)!}{2\cdot 4\cdot \ldots (2n-2)}=\frac{(2n-1)!}{(n-1)!2^{n-1}}$$

$$\prod_{i=1}^n (i+1)= 2\cdot 3\cdot 4 \ldots (n+1)=(n+1)!$$

$$\prod_{i=1}^n (zi+x)= (z+x)\cdot(2z+x)\cdot \ldots \cdot(nz+x)$$

The factorial notation $n!$ multiply every positive integers up to $n$.

Answer 2

The equation

$$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) = (2n-1)! $$

is (usually) wrong because, by definition,

$$ (2n-1)! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2n-1)$$

and we (usually) have

$$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) \neq 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2n-1)$$

The specific example you give, incidentally, comes up often enough to be given a name: the double factorial. It has different definitions depending on whether it is even or odd:

$$ (2n-1)!! = 1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1) $$ $$ (2n)!! = 2 \cdot 4 \cdot 6 \cdot \ldots \cdot (2n) $$

There are convenient identities

$$ n! = n!! \cdot (n-1)!!$$ $$ (2n)!! = 2^n n!$$

Answer 3

"but I still do not understand what is mathematically wrong with doing it like that."

Seriously?

You don't see why $1*2*3*4*5*6*7* ......(2n-3)*(2n-2)*(2n-1)*(2n)*(2n + 1) \ne 1*3*5*7*.......... *(2n-3)*(2n-1)*(2n+1)$?

Only one of those can be written as $(2n+1)!$. Which one is it going to be?

Oh, you chose the RHS...? Well, I wasn't actually giving you a choice. The question was first asked 80 years ago and everyone chose the LHS.

"However, my professor said that this was incorrect and that she had an explanation for why, but for some reason, it slipped her mind."

Seriously?!?!?