# Series of Multiplying Odd Numbers Involving Factorials?

• Jun 9th 2009, 04:06 PM
summermagic
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!
• Jun 9th 2009, 04:42 PM
Random Variable
n!! is by defintion equal to n*(n-2)*(n-4)...4*2 when n is a even positive integer, n*(n-2)*(n-4)...3*1 when n is an odd positive integer, and equals 1 if n=-1 or 0

so 1 +1*3 + 1*3*5 + ... = $\displaystyle \sum_{n \ odd} n!!$

maybe someone else knows a more interesting way to state the sum
• Jun 9th 2009, 04:50 PM
Chris L T521
Quote:

Originally Posted by summermagic
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!

That is not correct. Try your equation for $\displaystyle n=2$: $\displaystyle \sum_{n=1}^2\left(2n-1\right)!=1!+3!=1+6= 7\neq 4= 1+ 1*3$

Note that the pattern given is the same as:

$\displaystyle 1+\frac{1*2*3}{2}+\frac{1*2*3*4*5}{2*4}+\frac{1*2* 3*4*5*6*7}{2*4*6}+\dots$ $\displaystyle =1+\frac{3!}{2}+\frac{5!}{8}+\frac{7!}{48}+\dots$ $\displaystyle =\frac{1!}{2^0(0!)}+\frac{3!}{2^1(1!)}+\frac{5!}{2 ^2(2!)}+\frac{7!}{2^3(3!)}+\dots$

Thus, we have the following:

$\displaystyle 1+1*3+1*3*5+1*3*5*7+\dots = \sum_{n=1}^{\infty}\frac{\left(2n-1\right)!}{2^{n-1}\left(n-1\right)!}$

Does this make sense?

(Trust me, it took me a bit to notice the pattern.. XD)
• Jun 9th 2009, 06:19 PM
summermagic
Yes, it makes perfect sense now! Thank you so much!!! (Clapping)
• Oct 12th 2012, 09:29 PM
fiz
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