# Thread: Series of Multiplying Odd Numbers Involving Factorials?

1. ## 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: $(1)+(1*3)+(1*3*5)+(1*3*5*7)+(1*3*5*7*9)+...$

So basically it's the summation from $n=1$ to $\infty$ of plugging $n$ into an equation and then getting an odd number and then by plugging in every number less than $n$ 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:
$(2n-1)!$

But I'm not sure if that means
$(2n-1)(2(n-1)-1)(2(n-2)-1)...$ which would result in something like $9*7*5*3*1$ if I plug in $n=5$.

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

Sooo...if anyone actually read all of this and can tell me whether my final equation of $\sum_{n=1}^{\infty} (2n-1)!$ 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! Seriously though...thank you so much to whoever helps!

2. 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 + ... = $\sum_{n \ odd} n!!$

maybe someone else knows a more interesting way to state the sum

3. 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: $(1)+(1*3)+(1*3*5)+(1*3*5*7)+(1*3*5*7*9)+...$

So basically it's the summation from $n=1$ to $\infty$ of plugging $n$ into an equation and then getting an odd number and then by plugging in every number less than $n$ 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:
$(2n-1)!$

But I'm not sure if that means
$(2n-1)(2(n-1)-1)(2(n-2)-1)...$ which would result in something like $9*7*5*3*1$ if I plug in $n=5$.

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

Sooo...if anyone actually read all of this and can tell me whether my final equation of $\sum_{n=1}^{\infty} (2n-1)!$ 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! Seriously though...thank you so much to whoever helps!
That is not correct. Try your equation for $n=2$: $\sum_{n=1}^2\left(2n-1\right)!=1!+3!=1+6= 7\neq 4= 1+ 1*3$

Note that the pattern given is the same as:

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

Thus, we have the following:

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

Does this make sense?

(Trust me, it took me a bit to notice the pattern.. XD)

4. Yes, it makes perfect sense now! Thank you so much!!!

5. ## 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

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# how to denote factorial of odd numbers

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