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Math Help - Series of Multiplying Odd Numbers Involving Factorials?

  1. #1
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    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!
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  2. #2
    Super Member Random Variable's Avatar
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    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
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by summermagic View Post
    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)
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  4. #4
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    Yes, it makes perfect sense now! Thank you so much!!!
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  5. #5
    fiz
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    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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