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Math Help - Series Convergence Question i.e. Ratio Test

  1. #1
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    Series Convergence Question i.e. Ratio Test

    I basically need help with the algebra involved in this problem. Specifically an explanation of how the ! and the 2^n are going to cancel out. Thank You.


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  2. #2
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    Quote Originally Posted by ihatemath View Post
    I basically need help with the algebra involved in this problem. Specifically an explanation of how the ! and the 2^n are going to cancel out. Thank You.


    ratio test ...

    \displaystyle \lim_{n \to \infty} \frac{(n+2)!}{2^{n+1}} \cdot \frac{2^n}{(n+1)!}

    note that \displaystyle \frac{(n+2)!}{(n+1)!} = n+2

    and the limit simplifies to ...

    \displaystyle \lim_{n \to \infty} \frac{n+2}{2}
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    Yes, I got that far and got stuck as to why the second step you posted = n+2. I don't understand how you got from point A to point B
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    Quote Originally Posted by ihatemath View Post
    Yes, I got that far and got stuck as to why the second step you posted = n+2.
    Because n! = n(n-1)! = n(n-1)(n-2)! = n(n-1)(n-2)(n-3)! etc.
    i.e.  n! = \prod_{k=0}^{n-1}(n-k). So (n+2)! = (n+2)(n+2-1) = (n+2)(n+1)!
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  5. #5
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    \displaystyle \frac{(n+2)!}{(n+1)!} = \frac{(n+2)(n+1)(n)(n-1)(n-2) ... (3)(2)(1)}{(n+1)(n)(n-1)(n-2) ... (3)(2)(1)}
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