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Math Help - Limiit of a sequence

  1. #1
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    Limiit of a sequence

    For the following sequence : ((-1)^n) / (n!), if the sequence converges find its limit.

    How can you mathematically show that the limit is 0, i can see from observation that as n increases the denominator gets bigger and the term gets closer to 0, but how can i show it mathematically.

    This is actually practice question homework 4 marks so i'm a bit apprehensive about just saying that the limit is 0 from observation without any mathematical proof.

    Also by applying d'alemberts ratio test i get 0, is the result of the d'alembert ratio test the limit?
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  2. #2
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    Re: Limiit of a sequence

    \dfrac{-1}{n!} \le \dfrac{(-1)^n}{n!} \le \dfrac{1}{n!}

    By the Squeeze Theorem, since \lim_{n \to \infty} \dfrac{-1}{n!} = \lim_{n\to \infty} \dfrac{1}{n!} = 0, you have \lim_{n \to \infty} \dfrac{(-1)^n}{n!} = 0.

    To show that \lim_{n \to \infty} \dfrac{-1}{n!} = 0 = \lim_{n \to \infty} \dfrac{1}{n!}, you can use an \varepsilon proof. Given \varepsilon>0, can you show that there exists N \in \Bbb{N} such that for all n\ge N, \left|\dfrac{-1}{n!} - 0 \right| < \varepsilon and \left|\dfrac{1}{n!} - 0 \right| < \varepsilon? First, let's use an induction argument to show that for all n\in \Bbb{N}, n!\ge 2^{n-1}.

    Base of induction: n=0. Then 0! = 1 \ge 2^{0-1} = \dfrac{1}{2} is true. n=1. Then 1! = 1 \ge 2^{1-1} = 1 is also true.

    Assume the proposition is true for some positive integer n (we already showed it is true for n=1, so this assumption is true). Then (n+1)! = (n+1)n!. For all n\ge 1, n+1\ge 2. So, (n+1)n! \ge 2n! \ge 2\cdot 2^{n-1} = 2^{(n+1)-1} (the second inequality came from application of the induction assumption). Hence, n! \ge 2^{n-1} for all n.

    Now that is proven, choose any positive integer N > 1-\log_2 \varepsilon. Then for any n\ge N, we have n > 1-\log_2 \varepsilon, which implies 2^n > \dfrac{2}{\varepsilon}. Dividing both sides by 2 gives n! \ge 2^{n-1}>\dfrac{1}{\varepsilon}. Taking the reciprocal of both sides flips the inequality: \dfrac{1}{n!} < \varepsilon. This is exactly what you wanted to prove.

    There are various other ways of proving it.
    Thanks from topsquark
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  3. #3
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    Re: Limiit of a sequence

    Thank you very much, your explanation was very good and helped me learn a lot. however for this question is there a simple way to do this as i have very limited space provided to do this question so i think that my teacher wanted me to do it very quickly in a simple way, as its only worth 3 marks in a 100 mark paper. So is there a shorter way of mathematically proving limit is 0?
    Last edited by StudentUni; July 30th 2014 at 12:59 PM.
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  4. #4
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    Re: Limiit of a sequence

    Given \varepsilon>0, choose any positive integer N>1-\log_2 \varepsilon. For any n\ge N, we have:

    \left|\dfrac{(-1)^n}{n!}-0\right| = \dfrac{1}{n!} \le \dfrac{1}{2^{n-1}} \le \dfrac{1}{2^{N-1}} < \dfrac{1}{2^{1-\log_2 \varepsilon-1}} = \varepsilon. That's really all you need. If there is a question about the fact that 2^{n-1} \le n!, your instructor can ask you to prove it, but it is obviously true as 2^{n-1} = \underbrace{2\cdots 2}_{n-1\text{ times}} while n! = \underbrace{n\cdot (n-1)\cdots 2\cdot 1}_{n\text{ terms}} is a product of n numbers, all but one of them are at least 2.
    Last edited by SlipEternal; July 30th 2014 at 02:04 PM.
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  5. #5
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    Re: Limiit of a sequence

    An even easier argument. n! \ge n for all n>0
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  6. #6
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    Re: Limiit of a sequence

    Ok, thank you very much for your help.
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