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Math Help - Explanation on this series please

  1. #1
    Junior Member Shadow236's Avatar
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    Explanation on this series please

    The series that I'm working with is: (Series, n=1 -> infinity) n!/en^2

    I decided that the ratio test would be my best bet, so I have come to the limit(n-> infinity) (n+1)/(e2n+1).
    I know that this limit is infinity/infinity at the moment, which is indeterminate. Usually what I would do is place every term over the highest
    n power, but I get confused with the e term in the denominator.

    My book says that this limit = 0, which would make it convergent, I just get mixed up in the middle.

    What is the right approach to figuring out this limit?

    Thanks.
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    Re: Explanation on this series please

    Quote Originally Posted by Shadow236 View Post
    I have come to the limit(n-> infinity) (n+1)/(e2n+1).
    In general, for any real a and b such that a > 1, \lim_{n\to\infty}\frac{n^b}{a^n}=0. For a discussion about this, see StackExchange. In particular, answer #3 starting from "we need only look at n/c^n" has a rigorous proof that n/c^n\to0 as n\to\infty when c > 1.

    The function f(n) = e^n describes a chain reaction while g(n) = n describes simple burning. I recommend playing with the graphs of e^n and n at large n to see just how much faster e^n grows.
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    Junior Member Shadow236's Avatar
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    Re: Explanation on this series please

    I think I see what you're saying. So, because the denominator get's larger than the numerator so much faster, the limit can be said that it is equal to 0?
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    Re: Explanation on this series please

    Quote Originally Posted by Shadow236 View Post
    So, because the denominator get's larger than the numerator so much faster, the limit can be said that it is equal to 0?
    Yes. Of course, saying that the denominator gets larger than the numerator so much faster is not a proof, but there is a rigorous proof of this fact. Perhaps the simplest proof is the one that uses L'Hopital's rule if you are allowed to do it.
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    Re: Explanation on this series please

    Quote Originally Posted by emakarov View Post
    Yes. Of course, saying that the denominator gets larger than the numerator so much faster is not a proof, but there is a rigorous proof of this fact. Perhaps the simplest proof is the one that uses L'Hopital's rule if you are allowed to do it.
    You cannot use l'hospital's rule, it requires that the numerator and denominator be 0 in the limit. Also the function is not continuous as it is only defined for natural numbers n.
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    Re: Explanation on this series please

    Quote Originally Posted by Shakarri View Post
    You cannot use l'hospital's rule, it requires that the numerator and denominator be 0 in the limit.
    Why do you think so?

    Quote Originally Posted by Shakarri View Post
    Also the function is not continuous as it is only defined for natural numbers n.
    If (x+1)/e^{2x+1}\to0 over real numbers, then it trivially follows that (n+1)/e^{2n+1}\to0 over natural numbers.
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    Re: Explanation on this series please

    Quote Originally Posted by emakarov View Post
    Why do you think so?
    Thats what I remember the lecturer saying
    Consider the case
    \lim_{x->1} \frac{x^2}{x}

    By L'Hospital's rule
    \lim_{x->1} \frac{x^2}{x}=\lim_{x->1} \frac{2x}{1}

    Then 1=2

    It may seem trivial that if it tends to zero for natural numbers then it tends to zero for real numbers but that does not remove the fact that a function is not differentiable at a point if it is not continuous at that point.
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    Re: Explanation on this series please

    Quote Originally Posted by Shakarri View Post
    Thats what I remember the lecturer saying
    Consider the case
    \lim_{x->1} \frac{x^2}{x}

    By L'Hospital's rule
    \lim_{x->1} \frac{x^2}{x}=\lim_{x->1} \frac{2x}{1}

    Then 1=2
    In this example, the limits of x^2 and x are non-zero numbers. L'Hopital's rule is indeed not applicable in such case. But it is applicable if the limits of the numerator and denominator are infinities, which are not numbers. See Wikipedia, MathWorld, PlanetMath and Springer Encyclopedia of Mathematics.

    Quote Originally Posted by Shakarri View Post
    It may seem trivial that if it tends to zero for natural numbers then it tends to zero for real numbers
    No, the converse implication is trivial; this implication is false.

    Quote Originally Posted by Shakarri View Post
    but that does not remove the fact that a function is not differentiable at a point if it is not continuous at that point.
    So we first prove convergence for differentiable functions x+1 and e^{2x+1} and then use it to prove convergence on natural numbers.
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    Re: Explanation on this series please

    Thanks. I now see that tending to +-infinity makes sense just by inverting a function that tends to zero.

    Quote Originally Posted by emakarov View Post
    No, the converse implication is trivial; this implication is false.
    That was a typo. But its not really trivial if you have to go through a further proof you mention. You really shouldn't split up my sentence when the first part relies on the second.

    I don't know how you can use it to prove convergence on the natural numbers but ok. You suggest that a proof involving L'hospital's rule would perhaps be simple so it looked like you didn't realise an extra step was required.
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