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Math Help - ratio test for convergence

  1. #1
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    ratio test for convergence

    I'm doing some review of series and using the ratio test for convergence.
    I have the problem (the summation of) n!x^n/n^n
    I remember the ratio of convergence = e but I forgot the trick to get to there. Would someone mind refreshing me?
    Thanks!
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by morganfor View Post
    I'm doing some review of series and using the ratio test for convergence.
    I have the problem (the summation of) n!x^n/n^n
    I remember the ratio of convergence = e but I forgot the trick to get to there. Would someone mind refreshing me?
    Thanks!
    for a series \sum_{n=0}^{\infty} {a_n} you have,

    X = lim_(n \rightarrow \infty) |\frac{a_{n+1}}{a_n}|

    Ratio test will tell you that:

    if X<1, the series is absolute convergent
    if X>1, the series diverges
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by morganfor View Post
    I'm doing some review of series and using the ratio test for convergence.
    I have the problem (the summation of) n!x^n/n^n
    I remember the ratio of convergence = e but I forgot the trick to get to there. Would someone mind refreshing me?
    Thanks!
    Ratio test -> \lim_{n \to \infty} \bigg{|}\frac{a_{n+1}}{a_n}\bigg{|}

    So we get...

    \lim_{n \to \infty}\bigg{|}\frac{(n+1)!x^{n+1}n^n}{n! x^n (n+1)^{n+1}}\bigg{|}

    = \lim_{n \to \infty}\bigg{|}\frac{n!x^{n+1}n^n}{n! x^n (n+1)^n}\bigg{|}

    = \lim_{n \to \infty}\bigg{|}x\frac{n^n}{(n+1)^n}\bigg{|}

    = \lim_{n \to \infty}\bigg{|}x\bigg{(}\frac{n}{n+1}\bigg{)}^n\bi  gg{|}

    = \lim_{n \to  \infty}\bigg{|}x\bigg{(}1 + \frac{1}{n}\bigg{)}^n\bigg{|}


    = |x e^{-1}|
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