Series of Multiplying Odd Numbers Involving Factorials?

So for a bonus question, my teacher asked anyone if they could figure out how to write an equation for the following: $\displaystyle (1)+(1*3)+(1*3*5)+(1*3*5*7)+(1*3*5*7*9)+...$

So basically it's the summation from $\displaystyle n=1$ to $\displaystyle \infty$ of plugging $\displaystyle n$ into an equation and then getting an odd number and then by plugging in every number less than $\displaystyle n$ you will get the multiplication of an odd number and every odd number less than it until 1. Does that make sense? It's kind of like a factorial but only multiplying odd numbers.

Anyways, I think I kind of figured it out but I'm not too sure because I'm not really familiar with the rules of factorials, but here's what I came up with:

$\displaystyle (2n-1)!$

But I'm not sure if that means

$\displaystyle (2n-1)(2(n-1)-1)(2(n-2)-1)...$ which would result in something like $\displaystyle 9*7*5*3*1$ if I plug in $\displaystyle n=5$.

OR...if it means

$\displaystyle (2n-1)(2n-2)(2n-3)...$ which would result in something like $\displaystyle 9*8*7*6*5*4*3*2*1$ if $\displaystyle n=5$ was plugged in.

Sooo...if anyone actually read all of this and can tell me whether my final equation of $\displaystyle \sum_{n=1}^{\infty} (2n-1)!$ is correct or not and if not then how can I can fix it then I would be so appreciative and would give you a cookie...if I had one!(Giggle) Seriously though...thank you so much to whoever helps!

Re: Series of Multiplying Odd Numbers Involving Factorials?

Thank you Chris L. I was so close to figuring this out myself.. Sigh.. Thanks for your help