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Math Help - finding the limit of this function...

  1. #1
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    finding the limit of this function...

    After staring at this problem forever, I'm utterly at a loss as to how to approach this problem.

    the limit as n goes to infinity of (n^n)/[n!(2^n)]

    It has to do with whether the numerator or the denominator goes to infinity "faster" but I don't know how to start this problem off...

    Thanks in advance~
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  2. #2
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    Quote Originally Posted by choe83188 View Post
    After staring at this problem forever, I'm utterly at a loss as to how to approach this problem.

    the limit as n goes to infinity of (n^n)/[n!(2^n)]

    It has to do with whether the numerator or the denominator goes to infinity "faster" but I don't know how to start this problem off...

    Thanks in advance~
    use the ratio test to prove that the series \sum_{n=1}^{\infty} \frac{2^n n!}{n^n} converges and so \lim_{n\to\infty} \frac{2^n n!}{n^n} = 0. thus the limit of your sequence is \infty.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    use the ratio test to prove that the series \sum_{n=1}^{\infty} \frac{2^n n!}{n^n} converges and so \lim_{n\to\infty} \frac{2^n n!}{n^n} = 0. thus the limit of your sequence is \infty.

    Thanks much, hadn't thought of that. Will be doing that now.

    Thanks again!
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