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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.

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  • $\begingroup$ So for instance, is $1\times 3\times5\times 7$ equal to $7!$? Of course $7!=5040$, a nice even number. $\endgroup$
    – Angina Seng
    Mar 28, 2018 at 1:15
  • $\begingroup$ It's simply wrong. Make a substitution $2n-1=k$ then consider $k!$ and compare to your product $\endgroup$
    – Yuriy S
    Mar 28, 2018 at 1:15
  • $\begingroup$ If you want to say "although usually the expression $n! $ means the product of all natural numbers up to $n $, for the purpose of this paper, I'll be using it to mean 'go make a peanut butter sandwich'", you may. But you will have to expect no-one in any other context to know what you mean. $(zn+x)! $ simply does mean $1*2*...*(zn+x-1)*(zn+x) $ and it simply doesn't mean $(z+x)(2z+x)... (zn+x) $. But if you want to SAY it does, go ahead. But just make sure you explicitly state it, and you only use it in that one context. $\endgroup$
    – fleablood
    Mar 28, 2018 at 3:06
  • $\begingroup$ The OEIS is frequently a very useful resource for math. For this particular question, look at oeis.org/A001147 and oeis.org/A000165 $\endgroup$
    – Robert Soupe
    Mar 28, 2018 at 3:34

3 Answers 3

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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

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

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  • $\begingroup$ In case it's not clear, $!!$ is meant to be read as a single symbol. $n!!$ means to apply the double factorial operation to $n$; it does not mean to apply the factorial and then apply the factorial a second time. That is, we usually have $n!! \neq (n!)!$. $\endgroup$
    – user14972
    Mar 28, 2018 at 1:32
  • $\begingroup$ I did not know about the double factorial, thanks for letting me know. About that example I gave, that was a typo on my part. I will edit it to display what it should. $\endgroup$
    – Ii Meme
    Mar 29, 2018 at 2:12
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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 $n$.

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  • $\begingroup$ That's right, I forgot about product notation. I really appreciate your comment, I understand much better now $\endgroup$
    – Ii Meme
    Mar 29, 2018 at 2:01
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"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?!?!?

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  • $\begingroup$ Part of me thinks this may be a troll question. $\endgroup$
    – Crescendo
    Mar 28, 2018 at 4:47
  • $\begingroup$ Thanks for responding to my question. It was indeed serious; I legitimately did not know. Also, my tutor said that it was correct, I posted this question to figure out who was right. Have a nice day $\endgroup$
    – Ii Meme
    Mar 29, 2018 at 1:56

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