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Math Help - ratio test (series) urgent test tommorow!

  1. #1
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    ratio test (series) urgent test tommorow!

    i feel that the pic below explains my issue.

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  2. #2
    Senior Member vincisonfire's Avatar
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    n! means 1x2x3x...x(n-1)xn
    Therefore
    (n+1)! means 1x2x3x...x(n-1)xnx(n+1)
    So  \frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)}
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    Quote Originally Posted by vincisonfire View Post
    n! means 1x2x3x...x(n-1)xn
    Therefore
    (n+1)! means 1x2x3x...x(n-1)xnx(n+1)
    So  \frac{n!}{(n+1)!} = \frac{1x2x3x...x(n-1)xn}{x2x3x...x(n-1)xnx(n+1)} = \frac{1}{(n+1)}

    hm thanks kinda helps me but i m not fully understanding.. can any1 eleaborate?
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  4. #4
    Senior Member vincisonfire's Avatar
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    Well the definition on n! is the product of integers up to n.
    From this definition you can answer your question. I showed you that the product of integers up to n divided by the product of integers up to n+1 is equal to 1 over n+1.
    Then to compute the limit, you have to use l'Hospital rule. That is
     \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}
    Here  \lim_{x \rightarrow \infty} \frac{n+2}{(n+1)^2} = \lim_{x \rightarrow \infty} \frac{1}{2(n+1)} =0
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