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Math Help - Help verifying the series

  1. #1
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    Help verifying the series

    Hi, I'm not sure how to interpret this series so I was wondering if someone could clarify what it means.

    x^n/(1*3*5*...*(2n-1))

    Is the denominator just (2n-1)? Or do I separate the (2n-1) from the other numbers? If the latter, it would become (2n-1)^2 correct?
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  2. #2
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    Quote Originally Posted by My Little Pony View Post
    Hi, I'm not sure how to interpret this series so I was wondering if someone could clarify what it means.

    x^n/(1*3*5*...*(2n-1))

    Is the denominator just (2n-1)? Or do I separate the (2n-1) from the other numbers? If the latter, it would become (2n-1)^2 correct?
    Note that 1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!.

    So the series is \frac{x^n}{(2n - 1)!}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Note that 1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!.

    So the series is \frac{x^n}{(2n - 1)!}.
    Ah gotcha! Thanks a bunch!
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  4. #4
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    Do you need to work out if it converges? If so, the ratio test should do nicely...
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    I'm supposed to find the radius and interval of convergence. But it should be much easier now that I have that cleared up.
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    Quote Originally Posted by Prove It View Post
    Note that 1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!.

    So the series is \frac{x^n}{(2n - 1)!}.
    Are you sure this is correct?

    I thought (2n-1)! = (2n-1)(2n-2)(2n-3)...3x2x1?
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  7. #7
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    Quote Originally Posted by My Little Pony View Post
    Are you sure this is correct?

    I thought (2n-1)! = (2n-1)(2n-2)(2n-3)...3x2x1?
    Any odd number can be written as 2n - 1 if n \in \mathbf{N}.

    If n = 1 then 2n - 1 = 1.

    If n = 2 then 2n - 1 = 3.

    If n = 3 then 2n - 1 = 5. Etc.


    The counter is not 2n - 1. The counter is n.

    So, letting n \in \mathbf{N} we have

    (2n - 1)! = 1\times 3 \times 5 \times 7 \times \dots \times (2n - 5)\times(2n - 3)\times (2n - 1).
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    Quote Originally Posted by Prove It View Post
    Any odd number can be written as 2n - 1 if n \in \mathbf{N}.

    If n = 1 then 2n - 1 = 1.

    If n = 2 then 2n - 1 = 3.

    If n = 3 then 2n - 1 = 5. Etc.


    The counter is not 2n - 1. The counter is n.

    So, letting n \in \mathbf{N} we have

    (2n - 1)! = 1\times 3 \times 5 \times 7 \times \dots \times (2n - 5)\times(2n - 3)\times (2n - 1).
    My tutor informs me that what you're talking about is the double factorial of (2n-1).

    (2n-1)!!
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  9. #9
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    Quote Originally Posted by Prove It View Post
    Note that 1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = (2n - 1)!.

    [snip]
    No. In fact:

    1\times 3 \times 5 \times 7 \times \dots \times (2n - 1) = \frac{(2n)!}{n! 2^n}

    (easily proved by induction).
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