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Math Help - Proving limits

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    Proving limits

    Prove that (5^n)/n! approaches 0 as n approaches infinity.
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    Quote Originally Posted by Slazenger3 View Post
    Prove that (5^n)/n! approaches 0 as n approaches infinity.
    All you need to prove is that for large values of n that n! > 5^n.
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    Quote Originally Posted by Prove It View Post
    All you need to prove is that for large values of n that n! > 5^n.
    For large values of n we see that 2n>n. But, \lim\text{ }\frac{n}{2n}\ne 0
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    Quote Originally Posted by Drexel28 View Post
    For large values of n we see that 2n>n. But, \lim\text{ }\frac{n}{2n}\ne 0
    Yes but we're not proving it for \frac{n}{2n} are we?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    Yes but we're not proving it for \frac{n}{2n} are we?
    Yes. But proving that one is larger than the other is not sufficient to show that the limit is zero.
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    Quote Originally Posted by Drexel28 View Post
    Yes. But proving that one is larger than the other is not sufficient to show that the limit is zero.
    In this case it would be.

    I think it is bleedingly obvious that

    \frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}

    must go to 0 since the denominator is so much bigger than the numerator for large values of n.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    In this case it would be.

    I think it is bleedingly obvious that

    \frac{5 \cdot 5 \cdot 5 \dots 5}{n(n - 1)(n - 2)\dots 3 \cdot 2}

    must go to 0 since the denominator is so much bigger than the numerator for large values of n.
    Bleedingly obvious as this case may be. That does not prove the assertion that proving n!>5^n\implies \lim\text{ }\frac{5^n}{n!}=0
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    Quote Originally Posted by Drexel28 View Post
    Bleedingly obvious as this case may be. That does not prove the assertion that proving n!>5^n\implies \lim\text{ }\frac{5^n}{n!}=0
    A simple application of the ratio test shows that \displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n  (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0

    For reference, the ratio test says that \displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}}  {a_n}\bigg)
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    Quote Originally Posted by davismj View Post
    A simple application of the ratio test shows that \displaystyle\lim_{n\to\infty}\frac{5^n}{n!} = \displaystyle\lim_{n\to\infty}\frac{5^{n+1}n!}{5^n  (n+1)!} = \displaystyle\lim_{n\to\infty}\frac{5}{n} = 0

    For reference, the ratio test says that \displaystyle\lim_{n\to\infty} (a_n) = \displaystyle\lim_{n\to\infty}\bigg(\frac{a_{n+1}}  {a_n}\bigg)
    This is not a series. The ratio test only determines the convergence or divergence of a series. Not only that, it does not evaluate the series, it only tells us if the series is convergent or divergent.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    This is not a series. The ratio test only determines the convergence or divergence of a series. Not only that, it does not evaluate the series, it only tells us if the series is convergent or divergent.
    Yes. But he proved that \sum_{n=1}^{\infty}\frac{5^n}{n!} converges, and thus by necessity \frac{5^n}{n!}\to 0
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  11. #11
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    Quote Originally Posted by Drexel28 View Post
    Yes. But he proved that \sum_{n=1}^{\infty}\frac{5^n}{n!} converges, and thus by necessity \frac{5^n}{n!}\to 0
    Good point
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